If $\sin(\theta + \alpha) = a$ and $\sin(\theta + \beta) = b$, then prove that $\cos 2(\alpha - \beta) - 4ab\cos(\alpha - \beta) = 1 - 2a^2 - 2b^2$.

If sin(θ + α) = a and sin(θ + β) = b, then prove t – Class 11 Mathematics

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\sin(\theta + \alpha) = a

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Concept understanding — Trigonometric Identity Proof

Trigonometric Identity Proof: From Intuition to Precision

Imagine you're standing at the corner of a right triangle. The two shorter sides — one horizontal, one vertical — and the sloping hypotenuse are all connected. If you change the angle at your corner, the lengths of the sides change, but the relationship between them stays fixed. That fixed relationship is what a trigonometric identity captures.

The Core Idea

A trigonometric identity is an equation involving trigonometric functions (like $\sin \theta$ , $\cos \theta$ , $\tan \theta$ ) that is true for every angle $\theta$ where both sides are defined. It's not a conditional equation (like $\sin \theta = 0.5$ , which is true only for specific angles). It's an eternal truth about how these functions relate.

The most famous one is:

\sin^2 \theta + \cos^2 \theta = 1

This holds for any angle $\theta$ — acute, obtuse, negative, whatever. Why? Because on the unit circle, $\sin \theta$ is the $y$ -coordinate and $\cos \theta$ is the $x$ -coordinate of a point on a circle of radius 1. The Pythagorean theorem says $x^2 + y^2 = 1$ , so $\sin^2 \theta + \cos^2 \theta = 1$ is just the Pythagorean theorem in disguise.

Proving an Identity: The Method

When you're asked to prove a trigonometric identity, you're not solving for an angle. You're showing that the left-hand side (LHS) and right-hand side (RHS) are the same expression, just written differently.

The golden rule: Start with one side and transform it into the other, using known identities and algebraic manipulation. Never move terms across the equals sign as if solving an equation — that assumes the identity is already true, which is what you're trying to prove.

A Simple Example

Prove: $\tan \theta \cdot \cos \theta = \sin \theta$

Step 1: Pick a side to start with. Usually, the more complicated side is easier to simplify. Here, the LHS looks more complex.

Step 2: Replace $\tan \theta$ with $\frac{\sin \theta}{\cos \theta}$ (a known identity).

\tan \theta \cdot \cos \theta = \frac{\sin \theta}{\cos \theta} \cdot \cos \theta

Step 3: Cancel $\cos \theta$ (provided $\cos \theta \neq 0$ — but the identity holds for all angles where both sides are defined, and at $\cos \theta = 0$ , $\tan \theta$ is undefined anyway).

= \sin \theta

That's it. The LHS simplifies exactly to the RHS.

The Toolbox of Known Identities

To prove any identity, you need to know the basic building blocks:

Identity	Formula
Pythagorean	$\sin^2 \theta + \cos^2 \theta = 1$
Quotient	$\tan \theta = \frac{\sin \theta}{\cos \theta}$ , $\cot \theta = \frac{\cos \theta}{\sin \theta}$
Reciprocal	$\csc \theta = \frac{1}{\sin \theta}$ , $\sec \theta = \frac{1}{\cos \theta}$ , $\cot \theta = \frac{1}{\tan \theta}$
Even-Odd	$\sin(-\theta) = -\sin \theta$ , $\cos(-\theta) = \cos \theta$

Watch out

A common mistake is to treat $\sin^2 \theta$ as $(\sin \theta)^2$ — which it is — but then incorrectly think $\sin^2 \theta + \cos^2 \theta = 1$ means $\sin \theta + \cos \theta = 1$ . It does not. The square applies to the whole sine value, not to the angle.

A Slightly Harder Proof

Prove: $\frac{1 - \cos^2 \theta}{\cos \theta} = \sin \theta \tan \theta$

Start with LHS: $\frac{1 - \cos^2 \theta}{\cos \theta}$

From the Pythagorean identity, $1 - \cos^2 \theta = \sin^2 \theta$ . So:

\frac{\sin^2 \theta}{\cos \theta} = \sin \theta \cdot \frac{\sin \theta}{\cos \theta} = \sin \theta \tan \theta

That's the RHS. Done. …

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Concept: Trigonometric Identity Proof — expressing everything in terms of $\theta$ and the phase difference.

We are given:

\sin(\theta + \alpha) = a, \quad \sin(\theta + \beta) = b

Step 1: Write $\cos 2(\alpha - \beta)$ using the double-angle identity:

\cos 2(\alpha - \beta) = 1 - 2\sin^2(\alpha - \beta)

Step 2: Express $\sin(\alpha - \beta)$ using the given sines.

Let $u = \theta + \alpha$ , $v = \theta + \beta$ . Then $\alpha - \beta = u - v$ , so

\sin(\alpha - \beta) = \sin(u - v) = \sin u \cos v - \cos u \sin v

But $\sin u = a$ , $\sin v = b$ , and $\cos u = \pm\sqrt{1-a^2}$ , $\cos v = \pm\sqrt{1-b^2}$ .

However, the product $ab\cos(\alpha - \beta)$ will handle the signs.

Step 3: Compute the left-hand side directly using known expansions:

\cos 2(\alpha - \beta) = 1 - 2(\sin(\alpha - \beta))^2

and

\sin(\alpha - \beta) = \sin((\theta+\alpha) - (\theta+\beta)) = \sin(\theta+\alpha)\cos(\theta+\beta) - \cos(\theta+\alpha)\sin(\theta+\beta)

\sin(\alpha - \beta) = a\cos(\theta+\beta) - b\cos(\theta+\alpha)

Now square and use $\cos^2 = 1 - \sin^2$ :

$\sin^2(\alpha - \beta) = a^2(1-b^2) + b^2(1-a^2) - 2ab\cos(\theta+\alpha)\cos(\theta+\beta)$ …

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The key idea is to express $a$ and $b$ in terms of $\theta$ , then eliminate $\theta$ using trigonometric identities. The result is $\cos 2(\alpha - \beta) - 4ab\cos(\alpha - \beta) = 1 - 2a^2 - 2b^2$ .

We are given two equations involving $\theta$ , $\alpha$ , and $\beta$ . The goal is to prove a relation that involves only $\alpha$ and $\beta$ (and the constants $a$ , $b$ ), with $\theta$ completely eliminated. This is a classic elimination problem: we have two equations linking $\theta$ to $\alpha$ and $\beta$ , and we need to remove $\theta$ to get a pure identity in $\alpha - \beta$ .

The natural approach is to expand $\sin(\theta + \alpha)$ and $\sin(\theta + \beta)$ using the sine addition formula, then treat $\sin\theta$ and $\cos\theta$ as unknowns. We can solve for them in terms of $a$ , $b$ , $\alpha$ , $\beta$ , and then use the identity $\sin^2\theta + \cos^2\theta = 1$ to eliminate $\theta$ . That will yield the required relation.

Let's do it step by step.

Expand the given equations

\sin(\theta + \alpha) = \sin\theta \cos\alpha + \cos\theta \sin\alpha = a

\sin(\theta + \beta) = \sin\theta \cos\beta + \cos\theta \sin\beta = b

Think of these as two linear equations in the unknowns $x = \sin\theta$ and $y = \cos\theta$ .

Set up the system

\begin{cases} x \cos\alpha + y \sin\alpha = a \\ x \cos\beta + y \sin\beta = b \end{cases}

We can solve for $x$ and $y$ using determinants (Cramer's rule). The determinant of the coefficient matrix is:

D = \cos\alpha \sin\beta - \sin\alpha \cos\beta = \sin(\beta - \alpha) = -\sin(\alpha - \beta)

Solve for $x = \sin\theta$

x = \frac{\begin{vmatrix} a & \sin\alpha \\ b & \sin\beta \end{vmatrix}}{D} = \frac{a \sin\beta - b \sin\alpha}{\sin(\beta - \alpha)}

Similarly, for $y = \cos\theta$ :

y = \frac{\begin{vmatrix} \cos\alpha & a \\ \cos\beta & b \end{vmatrix}}{D} = \frac{b \cos\alpha - a \cos\beta}{\sin(\beta - \alpha)}

Use the Pythagorean identity Since $\sin^2\theta + \cos^2\theta = 1$ , we have:

\left( \frac{a \sin\beta - b \sin\alpha}{\sin(\beta - \alpha)} \right)^2 + \left( \frac{b \cos\alpha - a \cos\beta}{\sin(\beta - \alpha)} \right)^2 = 1

Multiply through by $\sin^2(\beta - \alpha)$ :

(a \sin\beta - b \sin\alpha)^2 + (b \cos\alpha - a \cos\beta)^2 = \sin^2(\beta - \alpha)

Expand both squares First term:

a^2 \sin^2\beta - 2ab \sin\alpha \sin\beta + b^2 \sin^2\alpha

Second term:

b^2 \cos^2\alpha - 2ab \cos\alpha \cos\beta + a^2 \cos^2\beta

Add them:

a^2(\sin^2\beta + \cos^2\beta) + b^2(\sin^2\alpha + \cos^2\alpha) - 2ab(\sin\alpha \sin\beta + \cos\alpha \cos\beta)

That simplifies to:

a^2 + b^2 - 2ab \cos(\alpha - \beta)

Because $\sin\alpha \sin\beta + \cos\alpha \cos\beta = \cos(\alpha - \beta)$ .

Right-hand side

$\sin^2(\beta - \alpha) = \sin^2(\alpha - \beta) = 1 - \cos^2(\alpha - \beta)$ .

So the equation becomes:

a^2 + b^2 - 2ab \cos(\alpha - \beta) = 1 - \cos^2(\alpha - \beta)

Rearrange to the desired form Bring all terms to one side: $\cos^2(\alpha - \beta) - 2ab \cos(\alpha - \beta) + a^2 + b^2 - 1 = 0$ …

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Q.If sin⁡(θ+α)=a\sin(\theta + \alpha) = asin(θ+α)=a and sin⁡(θ+β)=b\sin(\theta + \beta) = bsin(θ+β)=b, then prove that cos⁡2(α−β)−4abcos⁡(α−β)=1−2a2−2b2\cos 2(\alpha - \beta) - 4ab\cos(\alpha - \beta) = 1 - 2a^2 - 2b^2cos2(α−β)−4abcos(α−β)=1−2a2−2b2.

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Q.If $\sin(\theta + \alpha) = a$ and $\sin(\theta + \beta) = b$ , then prove that $\cos 2(\alpha - \beta) - 4ab\cos(\alpha - \beta) = 1 - 2a^2 - 2b^2$ .

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