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 sine functions.
We will use the following two formulas:
sin(a+b) = sin a cos b + cos a sin b …(i)
sin(a-b) = sin a cos b – cos a sin b …(ii)
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Formula of sin(a+b) sin(a-b)
sin(a+b) sin(a-b) Formula:
sin(a+b) sin(a-b) = $\sin^2 a -\sin^2 b$
Proof:
Using the above formulas (i) and (ii), we have
sin(a+b) sin(a-b)
= (sin a cos b + cos a sin b) (sin a cos b – cos a sin b)
= $(\sin a \cos b)^2$ $-(\cos a \sin b)^2$ by the formula $(x+y)(x-y)$ $=x^2-y^2$
= $\sin^2 a \cos^2 b -\cos^2 a \sin^2b$
= $\sin^2 a (1-\sin^2 b)$ $-(1-\sin^2 a) \sin^2b$ as we know that $\sin^2 \theta+\cos^2 \theta=1$
= sin2a – sin2a sin2b –sin2b + sin2a sin2b
= sin2 a -sin2 b
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In a similar way as above, we can prove the formula of sin(α+β) sin(α-β) = $\sin^2 \alpha -\sin^2 \beta$.
Formula of sin(α+β) sin(α-β)
Prove that sin(α+β) sin(α-β) = $\sin^2 \alpha -\sin^2 \beta$ $\cdots (\star)$
Proof:
Using the above formulas (i) and (ii), we have
sin(α+β) sin(α-β)
= (sin α cos β + cos α sin β) (sin α cos β – cos α sin β)
= $(\sin \alpha \cos \beta)^2$ $-(\cos \alpha \sin \beta)^2$ by the formula $(x+y)(x-y)$ $=x^2-y^2$
= $\sin^2 \alpha \cos^2 \beta -\cos^2 \alpha \sin^2 \beta$
= $\sin^2 \alpha (1-\sin^2 \beta)$ $-(1-\sin^2 \alpha) \sin^2 \beta$ using the formula of $\sin^2 \theta+\cos^2 \theta=1$
= $\sin^2 \alpha – \sin^2 \alpha \sin^2 \beta$ $-\sin^2 \beta +\sin^2 \alpha \sin^2 \beta$
= $\sin^2 \alpha -\sin^2 \beta$
Example 1: Find the value of $\sin 105^\circ \sin 15^\circ$
Solution:
Let α = $60^\circ$ and β = $45^\circ$in the above formula. So we have
$\sin 105^\circ \sin 15^\circ$
= $\sin(60^\circ+45^\circ) \sin(60^\circ-45^\circ)$
= $\sin^2 60^\circ – \sin^2 45^\circ$ by $(\star)$
= $(\dfrac{\sqrt{3}}{2})^2 – (\dfrac{1}{\sqrt{2}})^2$
= $\dfrac{3}{4} – \dfrac{1}{2}$
= $\dfrac{3-2}{4}$
= $\dfrac{1}{4}$
So the value of $\sin 105^\circ \sin 15^\circ$ is $\dfrac{1}{4}$.
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