Compound Angles: Build New Values from Known Ones

CBSE Mathematics · 2026–27 · Class 11

Learn with Rajesh Sir

You know 30° and 45°. Can you find 15° exactly? Compound-angle identities turn familiar angles into new values. Build the formulas, see what changes when an angle doubles, and learn to choose a useful transformation.

By the end, you can…
  • Derive and use sine/cosine sum and difference identities.
  • Apply tangent and cotangent formulas with their domain conditions.
  • Build double-angle and triple-angle identities.
  • Transform sums into products and products into sums.
  • Choose and justify a transformation without losing excluded inputs.
About 5 minutes
TOPIC 01

Derive the cosine of a difference

A compound angle is an angle written as a sum or difference, such as α+β\alpha+\beta or α−β\alpha-\beta. Its trigonometric value is generally not the sum or difference of the separate values. For example, cos⁡(60∘−30∘)=3/2\cos(60^\circ-30^\circ)=\sqrt3/2, whereas cos⁡60∘−cos⁡30∘=(1−3)/2\cos60^\circ-\cos30^\circ=(1-\sqrt3)/2.

To build the correct relationship, place points A=(cos⁡α,sin⁡α)A=(\cos\alpha,\sin\alpha) and B=(cos⁡β,sin⁡β)B=(\cos\beta,\sin\beta) on the unit circle. The distance formula gives

AB2=(cos⁡α−cos⁡β)2+(sin⁡α−sin⁡β)2AB^2=(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2=2−2(cos⁡αcos⁡β+sin⁡αsin⁡β).=2-2(\cos\alpha\cos\beta+\sin\alpha\sin\beta).

Now rotate both rays through −β-\beta. Rotation preserves their chord length. One point becomes (1,0)(1,0); the other has coordinates (cos⁡(α−β),sin⁡(α−β))(\cos(\alpha-\beta),\sin(\alpha-\beta)). Computing the same squared distance in this position gives 2−2cos⁡(α−β)2-2\cos(\alpha-\beta). Equate the two distance expressions to obtain

cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β.\boxed{\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta.}

The derivation uses only circle coordinates, the distance formula and the fact that rotation preserves distances. It does not assume the addition identity it is trying to prove. The unit circle also makes the result valid beyond acute angles.

Replace β\beta by −β-\beta, using cosine’s evenness and sine’s oddness. The difference formula then becomes the sum formula:

cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β.\boxed{\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.}

Read the signs carefully: cosine of a difference has a plus between the products; cosine of a sum has a minus. Move the two angles in the construction and compare the resulting chord and coordinates.

Work it through

Find an exact value.

cos⁡15∘=cos⁡(45∘−30∘)=2232+2212=6+24.\cos15^\circ=\cos(45^\circ-30^\circ)=\frac{\sqrt2}{2}\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\frac12=\frac{\sqrt6+\sqrt2}{4}.

Both products come from familiar exact angles; no decimal approximation is needed.

Explore this diagram with JavaScript enabled. The worked example above remains available to read and print.

Pause. Think. Try.

1. Which is the correct formula for cos⁡(A+B)\cos(A+B)?
A hint, pleaseThe cosine sum formula has a minus between its two products.
Read the answer and explanation

cos⁡Acos⁡B−sin⁡Asin⁡B\cos A\cos B-\sin A\sin B

Rotating the difference identity and using sine oddness gives cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B)=\cos A\cos B-\sin A\sin B.
2. What is cos⁡75∘\cos75^\circ?
A hint, pleaseWrite 75° as 45°+30°.
Read the answer and explanation

(6−2)/4(\sqrt6-\sqrt2)/4

cos⁡75∘=(2/2)(3/2)−(2/2)(1/2)=(6−2)/4\cos75^\circ=(\sqrt2/2)(\sqrt3/2)-(\sqrt2/2)(1/2)=(\sqrt6-\sqrt2)/4.
About 5 minutes
TOPIC 02

Let the sine formula follow from cosine

Sine and cosine exchange roles under a complementary angle: sin⁡t=cos⁡(π/2−t)\sin t=\cos(\pi/2-t). Combine that fact with the cosine difference formula rather than treating another pair of identities as unrelated facts to memorise.

Write

sin⁡(α+β)=cos⁡[(π2−α)−β].\sin(\alpha+\beta)=\cos\left[\left(\frac\pi2-\alpha\right)-\beta\right].

Apply cos⁡(U−V)=cos⁡Ucos⁡V+sin⁡Usin⁡V\cos(U-V)=\cos U\cos V+\sin U\sin V, with U=π/2−αU=\pi/2-\alpha and V=βV=\beta. Since cos⁡(π/2−α)=sin⁡α\cos(\pi/2-\alpha)=\sin\alpha and sin⁡(π/2−α)=cos⁡α\sin(\pi/2-\alpha)=\cos\alpha, this becomes

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β.\boxed{\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.}

Now replace β\beta by −β-\beta. Cosine stays unchanged and sine changes sign:

sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β.\boxed{\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.}

Sine therefore keeps the sum/difference sign between its products. In each product, one factor is a sine and the other a cosine. The arguments are paired in opposite orders. Saying these features aloud is a useful way to check a line before substituting values.

All four sine/cosine sum and difference identities are valid for every real pair of angles. You do not need to assume both angles are acute. For a quadrant problem, however, the individual sine and cosine values must carry their correct signs. A known magnitude such as 3/5 is not a signed coordinate until the quadrant is specified.

Use exact forms when the angles are familiar. For unfamiliar inputs, a decimal check can reveal a sign error, but the identity itself is an exact equality. Keep both arguments in the same angle unit throughout a calculation.

Work it through

Find two complementary exact values.

sin⁡75∘=sin⁡(45∘+30∘)=6+24,\sin75^\circ=\sin(45^\circ+30^\circ)=\frac{\sqrt6+\sqrt2}{4},sin⁡15∘=sin⁡(45∘−30∘)=6−24.\sin15^\circ=\sin(45^\circ-30^\circ)=\frac{\sqrt6-\sqrt2}{4}.

Notice that sin⁡75∘=cos⁡15∘\sin75^\circ=\cos15^\circ, as the complementary-angle relationship predicts.

Pause. Think. Try.

1. Expand sin⁡(A−B)\sin(A-B).
A hint, pleaseThe sine difference formula keeps the minus sign.
Read the answer and explanation

sin⁡Acos⁡B−cos⁡Asin⁡B\sin A\cos B-\cos A\sin B

Use the sine sum identity with input−B, then apply sin⁡(−B)=−sin⁡B\sin(-B)=-\sin B and cos⁡(−B)=cos⁡B\cos(-B)=\cos B.
2. What is sin⁡15∘\sin15^\circ?
A hint, pleaseUse 45°−30°.
Read the answer and explanation

(6−2)/4(\sqrt6-\sqrt2)/4

sin⁡15∘=(2/2)(3/2)−(2/2)(1/2)=(6−2)/4\sin15^\circ=(\sqrt2/2)(\sqrt3/2)-(\sqrt2/2)(1/2)=(\sqrt6-\sqrt2)/4.
About 5 minutes
TOPIC 03

Derive tangent and cotangent formulas with their conditions

Start with tan⁡(α+β)=sin⁡(α+β)/cos⁡(α+β)\tan(\alpha+\beta)=\sin(\alpha+\beta)/\cos(\alpha+\beta). Substitute the two compound-angle formulas. If cos⁡α≠0\cos\alpha\ne0 and cos⁡β≠0\cos\beta\ne0, divide numerator and denominator by cos⁡αcos⁡β\cos\alpha\cos\beta. This gives

tan⁡(α±β)=tan⁡α±tan⁡β1∓tan⁡αtan⁡β.\boxed{\tan(\alpha\pm\beta)=\frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}.}

The fraction also needs a nonzero final denominator. Equivalently, the individual tangents and the tangent of the resulting angle must all be defined. A formula involving tan⁡α\tan\alpha cannot be used when α=π/2\alpha=\pi/2, even if tan⁡(α−β)\tan(\alpha-\beta) happens to exist. In that situation return to sine and cosine or another valid angle split.

For cotangent, start with cosine divided by sine and divide the expanded terms by sin⁡αsin⁡β\sin\alpha\sin\beta. Where the individual and resulting cotangents are defined,

cot⁡(α+β)=cot⁡αcot⁡β−1cot⁡α+cot⁡β,\cot(\alpha+\beta)=\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta},cot⁡(α−β)=cot⁡αcot⁡β+1cot⁡β−cot⁡α.\cot(\alpha-\beta)=\frac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha}.

The order of the denominator in the difference formula matters. It follows directly from sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta, so it is cot⁡β−cot⁡α\cot\beta-\cot\alpha, not the reverse.

Conditions are part of the identity rather than a separate formality. They tell you whether the chosen algebraic route is permitted. Before plugging in values, check the two starting quotients and the final denominator. Then simplify exactly; rationalising a denominator can make the result easier to compare with another equivalent answer.

Work it through

Find tan⁡75∘\tan75^\circ.

tan⁡(45∘+30∘)=1+1/31−1/3=3+13−1=2+3.\tan(45^\circ+30^\circ)=\frac{1+1/\sqrt3}{1-1/\sqrt3}=\frac{\sqrt3+1}{\sqrt3-1}=2+\sqrt3.

The denominator is nonzero. The value is positive and greater than 1, matching a tangent input between 45° and 90°.

Pause. Think. Try.

1. Find tan⁡15∘\tan15^\circ.
A hint, pleaseUse 45°−30° in the tangent difference formula.
Read the answer and explanation

2−32-\sqrt3

(1−1/3)/(1+1/3)=(3−1)/(3+1)=2−3(1-1/\sqrt3)/(1+1/\sqrt3)=(\sqrt3-1)/(\sqrt3+1)=2-\sqrt3.
2. Why cannot the tangent sum fraction be evaluated with α=π/2\alpha=\pi/2?
A hint, pleaseCheck each quantity used in the displayed fraction.
Read the answer and explanation

tan⁡α\tan\alpha is undefined.

The derived quotient uses tan⁡α\tan\alpha, which divides by cos⁡(π/2)=0\cos(\pi/2)=0. Another valid identity may still evaluate the final angle.
About 5 minutes
TOPIC 04

Double an angle by setting the two inputs equal

A double-angle formula is a compound-angle identity with the same angle used twice. Setting α=β=x\alpha=\beta=x in the sine sum formula gives

sin⁡2x=2sin⁡xcos⁡x.\boxed{\sin2x=2\sin x\cos x.}

Doing the same in the cosine sum formula gives cos⁡2x=cos⁡2x−sin⁡2x\cos2x=\cos^2x-\sin^2x. Replace either square using sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1 to obtain three equivalent forms:

cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x.\boxed{\cos2x=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x.}

These forms serve different purposes. If a problem supplies cosine, 2cos⁡2x−12\cos^2x-1 may be quickest. If it supplies sine, choose 1−2sin⁡2x1-2\sin^2x. When a calculation contains a difference of squares, the first form may expose the structure immediately.

The tangent sum identity similarly gives

tan⁡2x=2tan⁡x1−tan⁡2x,\tan2x=\frac{2\tan x}{1-\tan^2x},

provided tan⁡x\tan x and tan⁡2x\tan2x are defined. In this fraction, 1−tan⁡2x≠01-\tan^2x\ne0. The sine/cosine forms themselves remain valid for every real xx.

Rearrange the cosine identities to reduce squares:

sin⁡2x=1−cos⁡2x2,cos⁡2x=1+cos⁡2x2.\sin^2x=\frac{1-\cos2x}{2},\qquad\cos^2x=\frac{1+\cos2x}{2}.

Notice that sin⁡2x\sin2x, 2sin⁡x2\sin x and sin⁡2x\sin^2x are three different instructions. The first doubles the angle, the second doubles the output and the third squares the output. A quick exact-angle test, such as x=30∘x=30^\circ, can prevent them from being confused.

When roots are introduced later, remember that sin⁡2x=∣sin⁡x∣\sqrt{\sin^2x}=|\sin x|. A square-root step must not silently discard the sign information that the circle supplies.

Work it through

Given sin⁡x=4/5\sin x=4/5, cos⁡x=3/5\cos x=3/5.

sin⁡2x=24535=2425,cos⁡2x=925−1625=−725.\sin2x=2\frac45\frac35=\frac{24}{25},\qquad\cos2x=\frac9{25}-\frac{16}{25}=-\frac7{25}.

Hence tan⁡2x=−24/7\tan2x=-24/7. The doubled angle lies in Quadrant II, consistent with positive sine and negative cosine.

Pause. Think. Try.

1. If sin⁡x=3/5\sin x=3/5 and cos⁡x=4/5\cos x=4/5, find sin⁡2x\sin2x.
A hint, pleaseUse 2 sinxcosx.
Read the answer and explanation

24/2524/25

2(3/5)(4/5)=24/252(3/5)(4/5)=24/25.
2. Which equals 1−cos⁡2x1-\cos2x for every real xx?
A hint, pleaseRearrange the form cos⁡2x=1−2sin⁡2x\cos2x=1-2\sin^2x.
Read the answer and explanation

2sin⁡2x2\sin^2x

Moving cos⁡2x\cos2x and 2sin⁡2x2\sin^2x to opposite sides gives the result.
About 5 minutes
TOPIC 05

Build triple angles from a double and a single angle

To calculate a triple angle, split 3x3x as 2x+x2x+x. The sine sum identity and double-angle formulas give

sin⁡3x=sin⁡2xcos⁡x+cos⁡2xsin⁡x\sin3x=\sin2x\cos x+\cos2x\sin x=2sin⁡xcos⁡2x+(1−2sin⁡2x)sin⁡x=2\sin x\cos^2x+(1-2\sin^2x)\sin xsin⁡3x=3sin⁡x−4sin⁡3x.\boxed{\sin3x=3\sin x-4\sin^3x.}

As a quick check, put x=0x=0: both sides of the sine identity are 0, while both sides of the cosine identity are 1. In the last step, replace cos⁡2x\cos^2x by 1−sin⁡2x1-\sin^2x and collect the sine terms. Deriving the result this way explains both the coefficient 3 and the minus sign before the cubic term.

For cosine, use cos⁡(2x+x)=cos⁡2xcos⁡x−sin⁡2xsin⁡x\cos(2x+x)=\cos2x\cos x-\sin2x\sin x. Replace cos⁡2x\cos2x by 2cos⁡2x−12\cos^2x-1, sin⁡2x\sin2x by 2sin⁡xcos⁡x2\sin x\cos x, and sin⁡2x\sin^2x by 1−cos⁡2x1-\cos^2x. Collecting terms yields

cos⁡3x=4cos⁡3x−3cos⁡x.\boxed{\cos3x=4\cos^3x-3\cos x.}

These two identities hold for every real angle. They also show why sin⁡3x\sin3x is not 3sin⁡x3\sin x: the cubic term is essential.

If t=tan⁡xt=\tan x, dividing the triple-angle sine and cosine expressions gives

tan⁡3x=3t−t31−3t2.\boxed{\tan3x=\frac{3t-t^3}{1-3t^2}.}

This version requires cos⁡x≠0\cos x\ne0 and 1−3t2≠01-3t^2\ne0, so that both tan⁡x\tan x and tan⁡3x\tan3x exist. Keep the sine/cosine derivation available when a tangent expression is undefined; do not repair a zero denominator by cancellation without checking the original expression.

For a numerical problem, choose the formula that matches the ratio actually given. If cosine alone is supplied, the cosine cubic avoids an unnecessary square root and the sign choice it would require.

Work it through

A tangent example. If tan⁡x=1/2\tan x=1/2, then

tan⁡3x=3/2−1/81−3/4=11/81/4=112.\tan3x=\frac{3/2-1/8}{1-3/4}=\frac{11/8}{1/4}=\frac{11}{2}.

The final denominator is 1/4, so the formula is valid. This calculation determines the tangent value without requiring a decimal approximation for xx.

Pause. Think. Try.

1. Which is the correct formula for sin⁡3x\sin3x?
A hint, pleaseExpand sin⁡(2x+x)\sin(2x+x).
Read the answer and explanation

3sin⁡x−4sin⁡3x3\sin x-4\sin^3x

Use the double-angle identities and cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x, then collect terms.
2. If cos⁡x=1/2\cos x=1/2, what is cos⁡3x\cos3x?
A hint, pleaseUse the cosine triple-angle cubic directly.
Read the answer and explanation

−1-1

4(1/2)3−3(1/2)=1/2−3/2=−14(1/2)^3-3(1/2)=1/2-3/2=-1.
About 5 minutes
TOPIC 06

Turn a sum into a product

Sometimes the useful feature is a pair of different angles added together. Introduce their average and half-difference:

u=A+B2,v=A−B2,A=u+v,B=u−v.u=\frac{A+B}{2},\qquad v=\frac{A-B}{2},\qquad A=u+v,\quad B=u-v.

Now apply the compound-angle formulas to sin⁡(u+v)+sin⁡(u−v)\sin(u+v)+\sin(u-v). The cos⁡usin⁡v\cos u\sin v terms cancel, leaving 2sin⁡ucos⁡v2\sin u\cos v. Replacing uu and vv gives the first of four sum-to-product identities:

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2,\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2},sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2,\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2,\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2.\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}.

The last minus sign is easy to miss. It follows from subtracting the cosine expansions, whose sine-product terms have opposite signs. It can also be checked with A=πA=\pi, B=0B=0: the left side is −2, so the right side must be negative.

Use these identities when averaging the two arguments makes a familiar angle or when a product will reveal a common factor. The original angles do not need to be acute, and there are no denominator restrictions in these four sine/cosine identities. If you later divide by a factor, that new step must still be checked.

Read the two new angles separately. The average (A+B)/2(A+B)/2 and half-difference (A−B)/2(A-B)/2 are not the original sum and difference. Writing them before selecting the functions reduces arithmetic and sign errors.

Work it through

A sum of exact values.

sin⁡75∘+sin⁡15∘=2sin⁡45∘cos⁡30∘=62.\sin75^\circ+\sin15^\circ=2\sin45^\circ\cos30^\circ=\frac{\sqrt6}{2}.

For an algebraic pair, cos⁡5x+cos⁡3x=2cos⁡4xcos⁡x\cos5x+\cos3x=2\cos4x\cos x. The average is 4x4x and the half-difference is xx.

Pause. Think. Try.

1. Transform cos⁡5x+cos⁡3x\cos5x+\cos3x into a product.
A hint, pleaseFind the average and half-difference of 5 x and 3 x.
Read the answer and explanation

2cos⁡4xcos⁡x2\cos4x\cos x

The average is 4x4x and the half-difference is xx, so use the cosine-sum identity.
2. Which equals cos⁡A−cos⁡B\cos A-\cos B?
A hint, pleaseThe cosine-difference identity has a minus in front.
Read the answer and explanation

−2sin⁡((A+B)/2)sin⁡((A−B)/2)-2\sin((A+B)/2)\sin((A-B)/2)

Subtract the expansions of cos⁡(u+v)\cos(u+v) and cos⁡(u−v)\cos(u-v); the cosine products cancel and the two negative sine products add.
About 5 minutes
TOPIC 07

Turn a product into a sum

The same compound-angle identities can be read in the opposite direction. Add sin⁡(A+B)\sin(A+B) and sin⁡(A−B)\sin(A-B): the cos⁡Asin⁡B\cos A\sin B terms cancel, leaving 2sin⁡Acos⁡B2\sin A\cos B. Subtract instead, and the other mixed product remains. Applying the same idea to the cosine pair gives four product-to-sum relationships:

2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B),2\sin A\cos B=\sin(A+B)+\sin(A-B),2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B),2\cos A\sin B=\sin(A+B)-\sin(A-B),2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B),2\cos A\cos B=\cos(A+B)+\cos(A-B),2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B).2\sin A\sin B=\cos(A-B)-\cos(A+B).

All four hold for real AA and BB. There is no division by a variable expression in their derivation. If your starting product lacks the factor 2, multiply and divide by 2—or simply use half the corresponding right side. Keep that numerical factor visible until the transformation is complete.

The order in the last formula is especially useful to check: cosine of the difference comes first, then minus cosine of the sum. For example, putting A=B=xA=B=x gives 2sin⁡2x=1−cos⁡2x2\sin^2x=1-\cos2x, exactly the double-angle rearrangement already established. This is a quick consistency check that also links the identity families.

Choose the direction that makes the next step simpler. A product may become a manageable sum of familiar angles; a sum may become a factor that cancels with a known nonzero denominator. There is no rule that every expression must end as a product or that every product should be expanded. The aim is an equivalent expression with a clear next step, while preserving every condition attached to the original calculation.

Work it through

Transform one product.

sin⁡3xsin⁡x=cos⁡2x−cos⁡4x2.\sin3x\sin x=\frac{\cos2x-\cos4x}{2}.

Here the difference of arguments is 2 x and their sum is 4 x. Also, 2sin⁡5xcos⁡2x=sin⁡7x+sin⁡3x2\sin5x\cos2x=\sin7x+\sin3x. Check that the angle arithmetic and the leading factor 2 agree with the chosen identity.

Pause. Think. Try.

1. Which equals 2sin⁡5xcos⁡2x2\sin5x\cos2x?
A hint, pleaseThe mixed product sin⁡Acos⁡B\sin A\cos B gives the sum of two sines.
Read the answer and explanation

sin⁡7x+sin⁡3x\sin7x+\sin3x

Use A=5xA=5x and B=2xB=2x in 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B=\sin(A+B)+\sin(A-B).
2. Transform sin⁡3xsin⁡x\sin3x\sin x.
A hint, pleaseUse half the sine-product identity.
Read the answer and explanation

(cos⁡2x−cos⁡4x)/2(\cos2x-\cos4x)/2

2sin⁡3xsin⁡x=cos⁡(3x−x)−cos⁡(3x+x)=cos⁡2x−cos⁡4x2\sin3x\sin x=\cos(3x-x)-\cos(3x+x)=\cos2x-\cos4x.
About 5 minutes
TOPIC 08

Choose an identity and keep the domain

Before expanding, look at the angle arguments. Are two angles being added? Is one twice another? Do two terms have a simple average? An identity becomes easier to choose when you see that structure before focusing on the coefficients. Then write the matching formula with its arguments clearly identified.

An identity is true for every input where its displayed expressions are defined. It is different from an equation that asks which particular inputs make an equality true. To prove an identity, transform one side into the other using valid equalities. You can also transform both sides into the same expression, provided you do not assume the desired equality as a starting fact.

Consider

sin⁡3x+sin⁡xcos⁡3x+cos⁡x.\frac{\sin3x+\sin x}{\cos3x+\cos x}.

Both sums have average 2 x and half-differencex. Sum-to-product gives

2sin⁡2xcos⁡x2cos⁡2xcos⁡x=tan⁡2x.\frac{2\sin2x\cos x}{2\cos2x\cos x}=\tan2x.

The original denominator must be nonzero. Since it equals 2cos⁡2xcos⁡x2\cos2x\cos x, that condition ensures both factors are nonzero and justifies the cancellation. The simplified tangent may exist at some additional inputs, but those inputs do not become valid for the original fraction. For instance, at x=π/2x=\pi/2 the original fraction is 0/0 even though tan⁡2x=0\tan2x=0.

For exact-value questions, choose an angle split made from values you know. For proof questions, choose a form that resembles the target. After simplifying, check one ordinary allowed input to catch arithmetic or sign mistakes. Such a check is useful evidence about your working, but it does not replace an algebraic proof valid over the whole stated domain.

Work it through

Apply the domain condition. At x=π/6x=\pi/6, the original denominator is cos⁡(π/2)+cos⁡(π/6)=3/2≠0\cos(\pi/2)+\cos(\pi/6)=\sqrt3/2\ne0. The fraction simplifies to tan⁡(π/3)=3\tan(\pi/3)=\sqrt3. At x=π/2x=\pi/2, it must remain undefined; the simplified expression cannot restore a missing original input.

Pause. Think. Try.

1. The fraction (sin⁡3x+sin⁡x)/(cos⁡3x+cos⁡x)(\sin3x+\sin x)/(\cos3x+\cos x) simplifies to which expression on its original domain?
A hint, pleaseUse sum-to-product in both numerator and denominator.
Read the answer and explanation

tan⁡2x\tan2x

The common factor 2cos⁡x2\cos x cancels where the original denominator is nonzero, leaving sin⁡2x/cos⁡2x\sin2x/\cos2x.
2. After cancelling a factor, what must happen to an input excluded by the original expression?
A hint, pleaseA simplification preserves values on the original domain; it does not redefine it.
Read the answer and explanation

It remains excluded.

An equivalent simplified formula may have a larger standalone domain. Equality with the original still carries the original exclusions.
PUT IT TOGETHER

Your chapter checkpoint

Mix the ideas. Try each question before opening its explanation.

1. Find sin⁡(45∘+30∘)\sin(45^\circ+30^\circ).
A hint, pleaseExpand using the sine addition formula.
Read the answer and explanation

(6+2)/4(\sqrt6+\sqrt2)/4

(2/2)(3/2)+(2/2)(1/2)=(6+2)/4(\sqrt2/2)(\sqrt3/2)+(\sqrt2/2)(1/2)=(\sqrt6+\sqrt2)/4.
2. If tan⁡x=1/2\tan x=1/2, find tan⁡2x\tan2x.
A hint, pleaseUse 2t/(1−t2)2t/(1-t^2).
Read the answer and explanation

4/34/3

tan⁡2x=1/(1−1/4)=4/3\tan2x=1/(1-1/4)=4/3. The denominator is nonzero.
3. Which equals sin⁡5x−sin⁡x\sin5x-\sin x?
A hint, pleaseThe average is 3 x and the half-difference is 2 x.
Read the answer and explanation

2cos⁡3xsin⁡2x2\cos3x\sin2x

Use sin⁡A−sin⁡B=2cos⁡((A+B)/2)sin⁡((A−B)/2)\sin A-\sin B=2\cos((A+B)/2)\sin((A-B)/2).
4. Which identity holds for all real xx, including where tangent is undefined?
A hint, pleaseChoose an equality involving only sine/cosine, which exist for every real angle.
Read the answer and explanation

cos⁡3x=4cos⁡3x−3cos⁡x\cos3x=4\cos^3x-3\cos x

The cosine triple-angle identity has no zero-denominator restriction. The quotient identities do.

Write the reasoning

Make a labelled sketch or a few lines of working. Then compare each step.

Use signed ratios in a compound angle

Angles AA and BB lie in QuadrantsII andIV respectively. Given sin⁡A=3/5\sin A=3/5 and cos⁡B=12/13\cos B=12/13, find cos⁡(A+B)\cos(A+B).
Compare your solution

Quadrant II gives cos⁡A=−4/5\cos A=-4/5. Quadrant IV gives sin⁡B=−5/13\sin B=-5/13. Therefore

cos⁡(A+B)=(−45)1213−35(−513)=−48+1565=−3365.\cos(A+B)=\left(-\frac45\right)\frac{12}{13}-\frac35\left(-\frac5{13}\right)=\frac{-48+15}{65}=-\frac{33}{65}.

Both sign choices came from the given quadrants, rather than from a positive square root alone.

Build a triple angle from one cosine

If cos⁡x=3/5\cos x=3/5, find cos⁡3x\cos3x without finding xx.
Compare your solution

The cosine cubic uses only the supplied cosine:

cos⁡3x=4(35)3−3(35)=108125−225125=−117125.\cos3x=4\left(\frac35\right)^3-3\left(\frac35\right)=\frac{108}{125}-\frac{225}{125}=-\frac{117}{125}.

No sine sign is needed. The answer lies between −1 and 1, as every cosine value must.

Prove a sum-to-product simplification

Show that (sin⁡5x−sin⁡x)/(cos⁡5x+cos⁡x)=tan⁡2x(\sin5x-\sin x)/(\cos5x+\cos x)=\tan2x wherever the original fraction is defined. State the domain condition used.
Compare your solution

Transform the numerator to 2cos⁡3xsin⁡2x2\cos3x\sin2x and the denominator to 2cos⁡3xcos⁡2x2\cos3x\cos2x. The original denominator condition is 2cos⁡3xcos⁡2x≠02\cos3x\cos2x\ne0, so both factors are nonzero and cancellation is valid. The quotient is sin⁡2x/cos⁡2x=tan⁡2x\sin2x/\cos2x=\tan2x. Retain both original restrictions cos⁡3x≠0\cos3x\ne0 and cos⁡2x≠0\cos2x\ne0, even where the standalone tangent formula might have a value.

Explore further · all the angles that work
Class 11 · classroom exploration

Every angle that works

Class 11 classroom exploration · formative. An inverse function returns one selected angle. A trigonometric equation can ask for every real angle with a given value. Use radians here, and let n∈Zn\in\mathbb Z represent any integer. The integer records the repeated turns; it is not a decimal parameter.

For sin⁡x=sin⁡a\sin x=\sin a, the unit circle gives two terminal-angle families: aa and π−a\pi-a, followed by complete turns. Therefore

x=2nπ+aorx=(2n+1)π−a.x=2n\pi+a\quad\text{or}\quad x=(2n+1)\pi-a.

The compact equivalent form is x=nπ+(−1)nax=n\pi+(-1)^n a. For sin⁡x=1/2\sin x=1/2, take a=π/6a=\pi/6: all solutions are x=π/6+2nπx=\pi/6+2n\pi or x=5π/6+2nπx=5\pi/6+2n\pi. Writing only sin⁡−1(1/2)=π/6\sin^{-1}(1/2)=\pi/6 would omit the other angles.

For cos⁡x=cos⁡a\cos x=\cos a, the two unit-circle positions are reflections in the horizontal axis. Thus

x=2nπ+aorx=2nπ−a,written as x=2nπ±a.x=2n\pi+a\quad\text{or}\quad x=2n\pi-a,\qquad\text{written as }x=2n\pi\pm a.

For example, cos⁡x=−1/2\cos x=-1/2 gives x=2nπ±2π/3x=2n\pi\pm2\pi/3. If a question restricts 0≤x<2π0\le x<2\pi, keep only 2π/32\pi/3 and 4π/34\pi/3. A general solution is a family; an interval answer is a selected list from that family.

Tangent repeats after a half-turn. If tan⁡a\tan a is defined, then

tan⁡x=tan⁡a⟹x=a+nπ.\tan x=\tan a\quad\Longrightarrow\quad x=a+n\pi.

Every angle in this family has the same nonzero cosine status as aa. For tan⁡x=1\tan x=1, the solutions are x=π/4+nπx=\pi/4+n\pi. Neither aa nor a candidate xx may be a tangent pole.

Check a solution family in two ways. Substitute a few integer values to test signs, then explain why the circle symmetry and period cover all positions. Numerical tests alone do not prove completeness. At special values such as sin⁡x=1\sin x=1, two displayed sine families can coincide; duplicate descriptions do not create new solutions.

When an equation contains sin⁡(2x)\sin(2x), first solve for the entire input 2x2x, then divide every term—including the repeated-turn term—by2. Finally filter against the original equation’s domain and any given interval. This keeps a principal angle, a complete solution family and an interval answer clearly separated.

For a complete example, solve cos⁡(2x)=−1/2\cos(2x)=-1/2. The cosine rule gives 2x=2nπ±2π/32x=2n\pi\pm2\pi/3, so x=nπ±π/3x=n\pi\pm\pi/3. On 0≤x<2π0\le x<2\pi, the answers are π/3,2π/3,4π/3,5π/3\pi/3,2\pi/3,4\pi/3,5\pi/3. Substitute each into the original doubled angle to check it. Notice that dividing the whole equation changed the repeated-turn term as well as the reference angle.

A concise written answer should name the period, give the complete family, state that the parameter is an integer, and show how any interval restriction selects the final angles.

Try the extension

1. For n∈Zn\in\mathbb Z, which gives all real solutions of sin⁡x=1/2\sin x=1/2?
A hint, pleaseFind the two positive angles in one complete turn, then repeat each by complete turns.
Read the answer and explanation

x=π/6+2nπx=\pi/6+2n\pi or x=5π/6+2nπx=5\pi/6+2n\pi

The unit-circle heights equal1/2 at π/6\pi/6 and 5π/65\pi/6. Both families repeat by 2π2\pi. A half-turn reverses the sine sign, so adding arbitrary multiples of π\pi to just one angle does not work.
2. For n∈Zn\in\mathbb Z, which gives all real solutions of tan⁡x=−3\tan x=-\sqrt3?
A hint, pleaseA principal angle is−π/3, and tangent repeats everyπ.
Read the answer and explanation

x=−π/3+nπx=-\pi/3+n\pi

tan⁡(−π/3)=−3\tan(-\pi/3)=-\sqrt3, and adding nπn\pi gives every tangent repetition. The two choices using only 2nπ2n\pi each omit alternate valid half-turns.
Class 11 milestone

Your Class 11 trigonometric-functions route is complete

You have connected all-angle definitions, six graphs and compound/multiple-angle identities. Revisit the graph checkpoint or begin Class 12 inverse trigonometry. For the Class 11 calculus connection, continue to trigonometric limits and derivatives, where the Class 11 and Class 12 steps are clearly separated.