Express sin3A in terms of sin A only. – Class 11 Business Mathematics Statistics | ConceptFirst

Worked Examples · Example 14

\sin3A

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Step 1 — write $3A=2A+A$ and expand. $\sin3A=\sin(2A+A)=\sin2A\cos A+\cos2A\sin A$ .

Step 2 — substitute the double-angle formulae. $\sin2A=2\sin A\cos A$ and $\cos2A=1-2\sin^2A$ :

\sin3A=(2\sin A\cos A)\cos A+(1-2\sin^2A)\sin A = 2\sin A\cos^2A+\sin A-2\sin^3A

Step 3 — eliminate $\cos A$ using $\cos^2A=1-\sin^2A$ . $2\sin A\cos^2A=2\sin A(1-\sin^2A)=2\sin A-2\sin^3A$ . Substituting back:

$\sin3A = (2\sin A-2\sin^3A)+\sin A-2\sin^3A = 3\sin A-4\sin^3A$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set MARCH1 markMCQ
$4\cos^3 40°-3\cos 40°$
(a) $\dfrac{1}{2}$
(b) $\dfrac{\sqrt{3}}{2}$
(c) $\dfrac{1}{\sqrt{2}}$
(d) $-\dfrac{1}{2}$
›Reveal solutionSolution

Recognise the triple-angle identity $4\cos^3\theta-3\cos\theta=\cos 3\theta$ ; here it gives $\cos 120°=-\dfrac{1}{2}$ .

The multiple-angle formula for cosine is

$\cos 3\theta=4\cos^3\theta-3\cos\theta.$

With $\theta=40°$ :

$4\cos^3 40°-3\cos 40°=\cos(3\times 40°)=\cos 120°.$

…

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CBSE 2023Set MARCH1 markMCQ
$\sin 15^\circ \cos 15^\circ$
(a) $\dfrac{\sqrt{3}}{2}$
(b) $1$
(c) $\dfrac{1}{4}$
(d) $\dfrac{1}{2}$
›Reveal solutionSolution

$\sin15^\circ\cos15^\circ = \tfrac12\sin30^\circ = \tfrac12\cdot\tfrac12 = \tfrac14$ , so the answer is option (c).

Recall the double-angle formula

$\sin 2\theta = 2\sin\theta\cos\theta \;\Rightarrow\; \sin\theta\cos\theta = \frac{1}{2}\sin 2\theta.$

With $\theta = 15^\circ$ :

…

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CBSE 2020Set MARCH1 markMCQ
$\sin A + \cos A = 1$
(a) $\dfrac{1}{2}$
(b) $1$
(c) $2$
(d) $0$
›Reveal solutionSolution

Squaring $\sin A+\cos A=1$ and applying $\sin^2A+\cos^2A=1$ leaves $\sin 2A=0$ .

Given:

$\sin A + \cos A = 1.$

Square both sides:

$(\sin A + \cos A)^2 = 1^2$

$\sin^2 A + 2\sin A\cos A + \cos^2 A = 1.$

Use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ : …

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