# Formula of cos(a+b) cos(a-b) | Formula of cos(α+β) cos(α-β)

In this post, we will establish the formula of sin(a+b) sin(a-b). Note that sin(a+b) sin(a-b) is a product of two cosine functions.

We will use the following two formulas:

cos(a+b) = cos a cos b – sin a sin b …(i)

cos(a-b) = cos a cos b + sin a sin b …(ii)

## Formula of cos(a+b) cos(a-b)

cos(a+b) cos(a-b) Formula: cos(a+b) cos(a-b) = $\cos^2 a -\sin^2 b$ = $\cos^2 b -\sin^2 a$

Proof:

Using the above formulas (i) and (ii), we have

cos(a+b) cos(a-b) = (cos a cos b – sin a sin b) (cos a cos b + sin a sin b)

= $(\cos a \cos b)^2$ $-(\sin a \sin b)^2$ Here we have used the formula (x+y)(x-y)=x2-y2

= $\cos^2 a \cos^2 b -\sin^2 a \sin^2b$ $\cdots (\star)$

= $\cos^2 a (1-\sin^2 b)$ $-(1-\cos^2 a) \sin^2b$ by the formula $sin^2 \theta+\cos^2\theta=1$

= $\cos^2 a – \cos^2 a \sin^2 b$ $-\sin^2b+\cos^2 a \sin^2b$

= cos2a -sin2b

So the formula of cos(a+b) cos(a-b) is cos2a -sin2b.

Next, we will prove that cos(a+b) cos(a-b) = cos2b -sin2a.

Proof:

From $(\star)$ we have that

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

= $(1-\sin^2 a) \cos^2 b$ $-\sin^2 a (1-\cos^2b)$

= $\cos^2 b – \sin^2 a \cos^2 b$ $-\sin^2 a+ \sin^2 a \cos^2b$

= cos2b -sin2a (Proved)

Formula of sin(a+b)sin(a-b)

Values of sin15, cos 15, tan 15

In a similar way as above, we can prove the formula of cos(α+β) cos(α-β) = cos2α -sin2β.

## Formula of cos(α+β) cos(α-β)

Prove that cos(α+β) cos(α-β) = cos2α -sin2β.

Proof:

Using the above formulas (i) and (ii), we have

sin(α+β) sin(α-β)

= (cos α cos β – sin α sin β) (cos α cos β + sin α sin β)

= $(\cos \alpha \cos \beta)^2$ $-(\sin \alpha \sin \beta)^2$ as we know (x+y)(x-y)=x2-y2

= $\cos^2 \alpha \cos^2 \beta -\sin^2 \alpha \sin^2 \beta$

= $\cos^2 \alpha (1-\sin^2 \beta)$ $-(1-\cos^2 \alpha) \sin^2 \beta$ by applying the formula of $\sin^2 \theta+\cos^2\theta=1$

= $\cos^2 \alpha – \cos^2 \alpha \sin^2 \beta$ $-\sin^2 \beta +\cos^2 \alpha \sin^2 \beta$

= cos2α -sin2β  (Proved)

Example 1: Find the value of $\cos 75^\circ \cos 15^\circ$

Solution:

Let α = $45^\circ$ and β = $30^\circ$ in the above formula. Thus we obtain that

$\cos 75^\circ \cos 15^\circ$

= $\cos(45^\circ+30^\circ) \cos(45^\circ-30^\circ)$

= $\cos^2 45^\circ – \sin^2 30^\circ$

= $(\dfrac{\sqrt{3}}{2})^2 – (\dfrac{1}{2})^2$

= $\dfrac{3}{4} – \dfrac{1}{4}$

= $\dfrac{3-1}{4}$

= $\dfrac{2}{4}$

= $\dfrac{1}{2}$

So the value of $\cos 75^\circ \cos 15^\circ$ is $\dfrac{1}{2}$.

## FAQs

Q1: What is the formula of cos(a+b)cos(a-b)?

Answer: The formula of cos(a+b)cos(a-b) is as follows: cos(a+b)cos(a-b) = cos2a -sin2b.