Exact Value of sin 75°

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

Write $75^\circ$ as a sum of special angles: $75^\circ = 45^\circ + 30^\circ$ = $A = 45^\circ$, $B = 30^\circ$

Apply the sine sum identity: $\sin 75^\circ = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$ = Two products of known values

Substitute the exact values: $\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$ = $\frac{\sqrt{6} + \sqrt{2}}{4}$

Check with a calculator: $\frac{\sqrt{6} + \sqrt{2}}{4} \approx \frac{2.4495 + 1.4142}{4} \approx 0.9659$ and $\sin 75^\circ \approx 0.9659$ = The values agree

Answer: $\sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.9659$

Exact Value of cos 15° with a Difference

Find the exact value of $\cos 15^\circ$.

Write $15^\circ$ as a difference of special angles: $15^\circ = 45^\circ - 30^\circ$ = $A = 45^\circ$, $B = 30^\circ$

Apply the cosine difference identity (the sign flips to $+$): $\cos 15^\circ = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$ = Plus sign for a difference

Substitute and simplify: $\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$ = $\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$

Notice the connection: $\cos 15^\circ = \sin 75^\circ$ because $15^\circ + 75^\circ = 90^\circ$ (cofunction identity) = Same value as in Example 1

Answer: $\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.9659$

Exact Value of tan 105°

Find the exact value of $\tan 105^\circ$ and simplify it.

Write $105^\circ$ as a sum: $105^\circ = 60^\circ + 45^\circ$, with $\tan 60^\circ = \sqrt{3}$ and $\tan 45^\circ = 1$ = $A = 60^\circ$, $B = 45^\circ$

Apply the tangent sum identity: $\tan 105^\circ = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}}$ = Denominator $1 - \sqrt{3}$ is negative

Rationalize: multiply top and bottom by $1 + \sqrt{3}$: $\frac{(\sqrt{3} + 1)^2}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2}$ = $-2 - \sqrt{3}$

Check the sign: $105^\circ$ is in Quadrant II, where tangent is negative; $-2 - \sqrt{3} \approx -3.7321$ = Negative, as expected

Answer: $\tan 105^\circ = -2 - \sqrt{3} \approx -3.7321$

Angles from Different Quadrants

$\sin A = \frac{3}{5}$ with $A$ in Quadrant II, and $\cos B = \frac{5}{13}$ with $B$ in Quadrant I. Find $\sin(A + B)$ and $\cos(A + B)$.

Find $\cos A$ with the Pythagorean identity: $\cos^2 A = 1 - \frac{9}{25} = \frac{16}{25}$, and cosine is negative in Quadrant II = $\cos A = -\frac{4}{5}$

Find $\sin B$: $\sin^2 B = 1 - \frac{25}{169} = \frac{144}{169}$, and sine is positive in Quadrant I = $\sin B = \frac{12}{13}$

Apply the sine sum identity: $\sin(A + B) = \frac{3}{5} \cdot \frac{5}{13} + \left(-\frac{4}{5}\right) \cdot \frac{12}{13} = \frac{15}{65} - \frac{48}{65}$ = $\sin(A + B) = -\frac{33}{65}$

Apply the cosine sum identity: $\cos(A + B) = \left(-\frac{4}{5}\right) \cdot \frac{5}{13} - \frac{3}{5} \cdot \frac{12}{13} = -\frac{20}{65} - \frac{36}{65}$ = $\cos(A + B) = -\frac{56}{65}$

Check with the Pythagorean identity: $\left(\frac{33}{65}\right)^2 + \left(\frac{56}{65}\right)^2 = \frac{1089 + 3136}{4225} = 1$ = The results fit together

Answer: $\sin(A + B) = -\frac{33}{65}$ and $\cos(A + B) = -\frac{56}{65}$, so $A + B$ lies in Quadrant III

Recognizing a Pattern Backwards

Simplify $\sin 70^\circ \cos 20^\circ + \cos 70^\circ \sin 20^\circ$ and $\cos 80^\circ \cos 20^\circ + \sin 80^\circ \sin 20^\circ$ without a calculator.

Match the first expression to a formula: $\sin A \cos B + \cos A \sin B$ with $A = 70^\circ$, $B = 20^\circ$ = It is $\sin(70^\circ + 20^\circ)$

Evaluate: $\sin 90^\circ = 1$ = First expression $= 1$

Match the second expression: $\cos A \cos B + \sin A \sin B$ (plus sign) is $\cos(A - B)$ = It is $\cos(80^\circ - 20^\circ)$

Evaluate: $\cos 60^\circ = \frac{1}{2}$ = Second expression $= \frac{1}{2}$

Answer: The first expression equals $1$, the second equals $\frac{1}{2}$

Mistake: Writing $\sin(A + B) = \sin A + \sin B$

Why: Sine is not a linear function, so you cannot split it over a sum. With $A = 30^\circ$ and $B = 60^\circ$: $\sin 90^\circ = 1$, but $\sin 30^\circ + \sin 60^\circ \approx 1.366$, which is not even a possible sine value.

Correct: $\sin(A + B) = \sin A \cos B + \cos A \sin B$. Move the sliders in the graph: the red curve $\sin x + \sin b$ leaves the other two.

Mistake: Keeping the sign in the cosine formula: $\cos(A + B) = \cos A \cos B + \sin A \sin B$

Why: Cosine flips the sign. Test it with $A = B = 45^\circ$: $\cos 90^\circ = 0$, but $\cos^2 45^\circ + \sin^2 45^\circ = 1$.

Correct: $\cos(A + B) = \cos A \cos B - \sin A \sin B$ and $\cos(A - B) = \cos A \cos B + \sin A \sin B$.

Mistake: Taking every missing value as positive, e.g. $\cos A = \frac{4}{5}$ when $\sin A = \frac{3}{5}$ and $A$ is in Quadrant II

Why: The Pythagorean identity only gives the size of the value. The quadrant decides the sign, and a wrong sign changes the whole result.

Correct: Use the quadrant: in Quadrant II cosine is negative, so $\cos A = -\frac{4}{5}$. Remember: Quadrant I all positive, II only sine, III only tangent, IV only cosine.

Mistake: Using $1 + \tan A \tan B$ in the denominator of $\tan(A + B)$

Why: In the tangent formula the numerator keeps the sign but the denominator flips it, just like in the cosine formula.

Correct: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

Combining Two Waves

In physics and electronics, a sum like $3\sin t + 4\cos t$ describes two overlapping waves. The sine sum identity shows that it is one single wave $R\sin(t + \varphi)$, and $R$ is its amplitude (how loud or how strong it is).

$R\sin(t + \varphi) = R\cos\varphi \sin t + R\sin\varphi \cos t$. Comparing with $3\sin t + 4\cos t$ gives $R\cos\varphi = 3$ and $R\sin\varphi = 4$.

Rotating Points in Games and Robotics

A point at distance $r$ and angle $\alpha$ has coordinates $x = r\cos\alpha$, $y = r\sin\alpha$. Turning it by $\beta$ gives the angle $\alpha + \beta$, and the sum identities turn that into the rotation formulas $x' = x\cos\beta - y\sin\beta$ and $y' = x\sin\beta + y\cos\beta$.

Rotating $(1, 0)$ by $90^\circ$: $x' = 1 \cdot 0 - 0 \cdot 1 = 0$ and $y' = 1 \cdot 1 + 0 \cdot 0 = 1$, so the point moves to $(0, 1)$.

The Angle Between Two Roads

A line with slope $m$ makes an angle $\theta$ with the $x$-axis where $\tan\theta = m$. The tangent difference identity gives the angle between two lines straight from their slopes, which planners use for road junctions, ramps and roof lines.

Lines with slopes $1$ and $0$: $\tan(\alpha - \beta) = \frac{1 - 0}{1 + 0} = 1$, so they meet at $45^\circ$.

Q: Why is there no simple rule like $\sin(A + B) = \sin A + \sin B$?

A: Because sine is not a linear function. Doubling an angle does not double its sine: $\sin 30^\circ = 0.5$, but $\sin 60^\circ \approx 0.866$, not $1$. The sum identity is the correct replacement for that wrong rule.

Q: How are the sum identities connected to the double angle identities?

A: Set $B = A$. Then $\sin(2A) = \sin A \cos A + \cos A \sin A = 2\sin A \cos A$ and $\cos(2A) = \cos^2 A - \sin^2 A$. The tangent formula gives $\tan(2A) = \frac{2\tan A}{1 - \tan^2 A}$.

Q: Which pair of special angles should I use?

A: Any pair works if it gives the angle: $75^\circ = 45^\circ + 30^\circ$, $15^\circ = 45^\circ - 30^\circ$ or $60^\circ - 45^\circ$, $105^\circ = 60^\circ + 45^\circ$, $165^\circ = 120^\circ + 45^\circ$ or $135^\circ + 30^\circ$. Different pairs lead to the same final value.

Q: When is $\tan(A + B)$ undefined?

A: When $1 - \tan A \tan B = 0$, that is when $\tan A \tan B = 1$. Then $A + B$ is $90^\circ$ (or $270^\circ$), where tangent does not exist. Example: $\tan 30^\circ \tan 60^\circ = 1$ and $30^\circ + 60^\circ = 90^\circ$.