AlgebraDifficulty 30-904 tagged problems

Trigonometric Identities

Trigonometric Identities is the habit of rewriting trigonometric expressions with Pythagorean, angle-addition, and double-angle identities. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$\sin^2\theta + \cos^2\theta = 1,\qquad \sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta,\qquad \cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta.$$

A small set of identities — the Pythagorean relation and the angle-addition formulas — lets you simplify, combine, or exactly evaluate trigonometric expressions that resist direct computation.

Why It Works

1
On the unit circle a point at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$; the circle equation $x^2 + y^2 = 1$ gives $\sin^2\theta + \cos^2\theta = 1$ immediately.
2
Rotating by $\alpha + \beta$ composes two rotations, which (via the rotation matrix or complex exponentials $e^{i(\alpha+\beta)} = e^{i\alpha}e^{i\beta}$) produces the angle-addition formulas for $\sin$ and $\cos$.
3
Setting $\alpha = \beta = \theta$ specializes them to the double-angle forms $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = \cos^2\theta - \sin^2\theta$.
4
To attack a problem, choose the identity that turns the unknown angle into known reference angles ($30^\circ, 45^\circ, 60^\circ$) or collapses a product/sum into something the Pythagorean identity can finish.

Worked Examples

Example 1

Find the exact value of $\cos 75^\circ$.

1
Write $75^\circ = 45^\circ + 30^\circ$ so the value comes from known reference angles.
2
Apply the addition formula: $\cos(45^\circ + 30^\circ) = \cos 45^\circ\cos 30^\circ - \sin 45^\circ\sin 30^\circ$.
3
Substitute exact values: $\frac{\sqrt 2}{2}\cdot\frac{\sqrt 3}{2} - \frac{\sqrt 2}{2}\cdot\frac{1}{2} = \frac{\sqrt 6}{4} - \frac{\sqrt 2}{4} = \frac{\sqrt 6 - \sqrt 2}{4}$.

Answer:

$\cos 75^\circ = \dfrac{\sqrt 6 - \sqrt 2}{4}$.

Warm-up

Find the exact value of $\sin 75^\circ$.

1
Write $75^\circ = 45^\circ + 30^\circ$ to use reference angles.
2
Apply the addition formula: $\sin(45^\circ + 30^\circ) = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ$.
3
Substitute exact values: $\dfrac{\sqrt 2}{2}\cdot\dfrac{\sqrt 3}{2} + \dfrac{\sqrt 2}{2}\cdot\dfrac{1}{2} = \dfrac{\sqrt 6}{4} + \dfrac{\sqrt 2}{4}$.

Answer:

$\sin 75^\circ = \dfrac{\sqrt 6 + \sqrt 2}{4}$.

Contest level

Given $\sin\theta + \cos\theta = \dfrac{1}{2}$, find $\sin^3\theta + \cos^3\theta$.

1
Square the given: $(\sin\theta + \cos\theta)^2 = \dfrac{1}{4}$, i.e. $\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = \dfrac{1}{4}$. Using the Pythagorean identity, $1 + 2\sin\theta\cos\theta = \dfrac{1}{4}$, so $\sin\theta\cos\theta = -\dfrac{3}{8}$.
2
Factor the sum of cubes: $\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta) = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$.
3
Substitute: $\dfrac{1}{2}\left(1 - \left(-\dfrac{3}{8}\right)\right) = \dfrac{1}{2}\cdot\dfrac{11}{8} = \dfrac{11}{16}$.

Answer:

$\dfrac{11}{16}$.

Olympiad / Challenge

Compute $\cos 20^\circ \cos 40^\circ \cos 80^\circ$.

1
Multiply by $\sin 20^\circ$ and use $2\sin A\cos A = \sin 2A$ repeatedly: $\sin 20^\circ\cos 20^\circ\cos 40^\circ\cos 80^\circ = \tfrac{1}{2}\sin 40^\circ\cos 40^\circ\cos 80^\circ$.
2
Continue: $\tfrac{1}{2}\sin 40^\circ\cos 40^\circ\cos 80^\circ = \tfrac{1}{4}\sin 80^\circ\cos 80^\circ = \tfrac{1}{8}\sin 160^\circ$.
3
Since $\sin 160^\circ = \sin 20^\circ$, the whole expression is $\sin 20^\circ\cdot(\cos 20^\circ\cos 40^\circ\cos 80^\circ) = \tfrac{1}{8}\sin 20^\circ$. Dividing by $\sin 20^\circ \ne 0$ gives the product $= \tfrac{1}{8}$.

Answer:

$\dfrac{1}{8}$.

Going Deeper

Generalization: the angle-addition formulas are the single source for double-angle, half-angle, sum-to-product, and product-to-sum identities; via $e^{i\theta} = \cos\theta + i\sin\theta$ they are just the statement $e^{i(\alpha+\beta)} = e^{i\alpha}e^{i\beta}$, and they generate the multiple-angle (Chebyshev) polynomials.
Where it appears: exact evaluation of non-standard angles (AMC), simplifying products of cosines via a telescoping $\sin$-multiplier (olympiad), and converting between $a\sin\theta + b\cos\theta$ and the single sinusoid $R\sin(\theta + \varphi)$.
Pitfall: the sign in the cosine addition formula is FLIPPED relative to sine: $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ (minus), while $\sin(\alpha + \beta)$ uses plus. Mixing the signs, or forgetting degrees-vs-radians, is the most common evaluation error.

Spot the Signal

Use it when an expression mixes $\sin$, $\cos$, or $\tan$ and a direct evaluation stalls.
You can describe the hard part as rewriting trigonometric expressions with Pythagorean, angle-addition, and double-angle identities, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by choosing the identity that collapses the messy combination into a single function.
Name the relation that makes Trigonometric Identities legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Reaching for an identity that doesn't match the angle structure, or dropping a sign in a double-angle step.
Applying Trigonometric Identities because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Pick the identity that removes a function, not the one you remember best.

Try it on:

Evaluate or simplify a contest trigonometric expression where the obstacle is choosing the right identity.

Back to techniques

AlgebraDifficulty 30-904 tagged problems

Trigonometric Identities

Practice this technique Match my level

The Key Result

$$\sin^2\theta + \cos^2\theta = 1,\qquad \sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta,\qquad \cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta.$$

A small set of identities — the Pythagorean relation and the angle-addition formulas — lets you simplify, combine, or exactly evaluate trigonometric expressions that resist direct computation.

Why It Works

1
On the unit circle a point at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$; the circle equation $x^2 + y^2 = 1$ gives $\sin^2\theta + \cos^2\theta = 1$ immediately.
2
Rotating by $\alpha + \beta$ composes two rotations, which (via the rotation matrix or complex exponentials $e^{i(\alpha+\beta)} = e^{i\alpha}e^{i\beta}$) produces the angle-addition formulas for $\sin$ and $\cos$.
3
Setting $\alpha = \beta = \theta$ specializes them to the double-angle forms $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = \cos^2\theta - \sin^2\theta$.
4
To attack a problem, choose the identity that turns the unknown angle into known reference angles ($30^\circ, 45^\circ, 60^\circ$) or collapses a product/sum into something the Pythagorean identity can finish.

Worked Examples

Example 1

Find the exact value of $\cos 75^\circ$.

1
Write $75^\circ = 45^\circ + 30^\circ$ so the value comes from known reference angles.
2
Apply the addition formula: $\cos(45^\circ + 30^\circ) = \cos 45^\circ\cos 30^\circ - \sin 45^\circ\sin 30^\circ$.
3
Substitute exact values: $\frac{\sqrt 2}{2}\cdot\frac{\sqrt 3}{2} - \frac{\sqrt 2}{2}\cdot\frac{1}{2} = \frac{\sqrt 6}{4} - \frac{\sqrt 2}{4} = \frac{\sqrt 6 - \sqrt 2}{4}$.

Answer:

$\cos 75^\circ = \dfrac{\sqrt 6 - \sqrt 2}{4}$.

Warm-up

Find the exact value of $\sin 75^\circ$.

1
Write $75^\circ = 45^\circ + 30^\circ$ to use reference angles.
2
Apply the addition formula: $\sin(45^\circ + 30^\circ) = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ$.
3
Substitute exact values: $\dfrac{\sqrt 2}{2}\cdot\dfrac{\sqrt 3}{2} + \dfrac{\sqrt 2}{2}\cdot\dfrac{1}{2} = \dfrac{\sqrt 6}{4} + \dfrac{\sqrt 2}{4}$.

Answer:

$\sin 75^\circ = \dfrac{\sqrt 6 + \sqrt 2}{4}$.

Contest level

Given $\sin\theta + \cos\theta = \dfrac{1}{2}$, find $\sin^3\theta + \cos^3\theta$.

1
Square the given: $(\sin\theta + \cos\theta)^2 = \dfrac{1}{4}$, i.e. $\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = \dfrac{1}{4}$. Using the Pythagorean identity, $1 + 2\sin\theta\cos\theta = \dfrac{1}{4}$, so $\sin\theta\cos\theta = -\dfrac{3}{8}$.
2
Factor the sum of cubes: $\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta) = (\sin\theta + \cos\theta)(1 - \sin\theta\cos\theta)$.
3
Substitute: $\dfrac{1}{2}\left(1 - \left(-\dfrac{3}{8}\right)\right) = \dfrac{1}{2}\cdot\dfrac{11}{8} = \dfrac{11}{16}$.

Answer:

$\dfrac{11}{16}$.

Olympiad / Challenge

Compute $\cos 20^\circ \cos 40^\circ \cos 80^\circ$.

1
Multiply by $\sin 20^\circ$ and use $2\sin A\cos A = \sin 2A$ repeatedly: $\sin 20^\circ\cos 20^\circ\cos 40^\circ\cos 80^\circ = \tfrac{1}{2}\sin 40^\circ\cos 40^\circ\cos 80^\circ$.
2
Continue: $\tfrac{1}{2}\sin 40^\circ\cos 40^\circ\cos 80^\circ = \tfrac{1}{4}\sin 80^\circ\cos 80^\circ = \tfrac{1}{8}\sin 160^\circ$.
3
Since $\sin 160^\circ = \sin 20^\circ$, the whole expression is $\sin 20^\circ\cdot(\cos 20^\circ\cos 40^\circ\cos 80^\circ) = \tfrac{1}{8}\sin 20^\circ$. Dividing by $\sin 20^\circ \ne 0$ gives the product $= \tfrac{1}{8}$.

Answer:

$\dfrac{1}{8}$.

Going Deeper

Generalization: the angle-addition formulas are the single source for double-angle, half-angle, sum-to-product, and product-to-sum identities; via $e^{i\theta} = \cos\theta + i\sin\theta$ they are just the statement $e^{i(\alpha+\beta)} = e^{i\alpha}e^{i\beta}$, and they generate the multiple-angle (Chebyshev) polynomials.
Where it appears: exact evaluation of non-standard angles (AMC), simplifying products of cosines via a telescoping $\sin$-multiplier (olympiad), and converting between $a\sin\theta + b\cos\theta$ and the single sinusoid $R\sin(\theta + \varphi)$.
Pitfall: the sign in the cosine addition formula is FLIPPED relative to sine: $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ (minus), while $\sin(\alpha + \beta)$ uses plus. Mixing the signs, or forgetting degrees-vs-radians, is the most common evaluation error.

Spot the Signal

Use it when an expression mixes $\sin$, $\cos$, or $\tan$ and a direct evaluation stalls.
You can describe the hard part as rewriting trigonometric expressions with Pythagorean, angle-addition, and double-angle identities, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by choosing the identity that collapses the messy combination into a single function.
Name the relation that makes Trigonometric Identities legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Reaching for an identity that doesn't match the angle structure, or dropping a sign in a double-angle step.
Applying Trigonometric Identities because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Pick the identity that removes a function, not the one you remember best.

Try it on:

Evaluate or simplify a contest trigonometric expression where the obstacle is choosing the right identity.