sin(45° + 30°) = sin45·cos30 + cos45·sin30.

Find sin 75° Using Compound Angle Formula

Question

Find the value of $\sin 75°$ .

Solution — Step by Step

We need $\sin 75°$ , but 75° is not a standard angle. The trick is to write it as a sum of two angles we do know: $75° = 45° + 30°$ .

Both 45° and 30° have exact values we can work with.

The formula for $\sin(A + B)$ is:

\sin(A + B) = \sin A \cos B + \cos A \sin B

Substitute $A = 45°$ , $B = 30°$ :

\sin 75° = \sin 45° \cos 30° + \cos 45° \sin 30°

We know these by heart (and you should too — they’re in every exam):

Angle	sin	cos
30°	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$
45°	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$

Substituting:

\sin 75° = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2}

\sin 75° = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}}

Rationalise the denominator by multiplying numerator and denominator by $\sqrt{2}$ :

\sin 75° = \frac{(\sqrt{3} + 1)\sqrt{2}}{2 \cdot 2} = \frac{\sqrt{6} + \sqrt{2}}{4}

Final Answer: $\sin 75° = \dfrac{\sqrt{6} + \sqrt{2}}{4}$

Why This Works

The compound angle formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$ comes directly from the unit circle and the geometry of projections. When we add two angles, the sine of the combined rotation picks up contributions from both components — which is exactly what the right-hand side captures.

The key insight is that we can only get exact values for angles like 30°, 45°, 60°, 90°. Any other angle needs to be built from these using sum, difference, or double-angle formulas. This is why the question says “using compound angle formula” — it’s a hint, not a suggestion.

This result ( $\frac{\sqrt{6}+\sqrt{2}}{4}$ ) also comes up when you check $\cos 15°$ . That’s not a coincidence — $\sin 75° = \cos 15°$ because they’re complementary angles. Good to remember as a self-check.

Alternative Method

We can also write $75° = 90° - 15°$ , which gives us $\sin 75° = \cos 15°$ .

Then find $\cos 15° = \cos(45° - 30°)$ using the formula:

\cos(A - B) = \cos A \cos B + \sin A \sin B

\cos 15° = \cos 45° \cos 30° + \sin 45° \sin 30° = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}

Same answer — good sign we haven’t made an error.

In JEE Main 2024, a similar question asked for $\sin 15°$ or $\cos 75°$ . The answer is $\frac{\sqrt{6}-\sqrt{2}}{4}$ — same structure but with a minus sign. The difference/sum of formulas give you both in one shot, so learn the pair together.

Common Mistake

The most frequent error: students write $\sin 75° = \sin 45° + \sin 30°$ . This is completely wrong — sine is not a linear function. $\sin(A + B) \neq \sin A + \sin B$ .

If you add them directly: $\frac{1}{\sqrt{2}} + \frac{1}{2} \approx 0.707 + 0.5 = 1.207$ . But sine can never exceed 1. That alone tells you something went wrong.

Always expand $\sin(A + B)$ using the full formula — never split the sine across a sum inside.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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