Question

If $cos(\theta -\alpha )$ = a, $sin(\theta -\beta )$ = b,

then $co{s}^2(\alpha -\beta )$ + 2ab $sin(\alpha -\beta )$ is equal to

• A. $4a^2{b}^2$
• B. $a^2-b^2$
• C. $a^2+b^2$
• D. -$a^2{b}^2$

Answer 1

The correct option is C

$a^2+b^2$

We have $sin(\alpha -\beta )$ = $sin(\theta -\beta -\overline{\theta -\alpha })$

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

= ba - $\sqrt{1-b^2}\sqrt{1-a^2}$

And $cos(\alpha -\beta )=cos(\theta -\beta -\overline{\theta -\alpha })$

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

= $a\sqrt{1-b^2}+b\sqrt{1-a^2}$

∴ Given expression is $co{s}^2(\alpha -\beta )+2absin(\alpha -\beta )$

= $(a\sqrt{1-b^2}+b\sqrt{1-a^2}{)}^2$ + 2ab{$ab-\sqrt{1-a^2}\sqrt{1-b^2}$}

= $a^2+b^2$.

Trick: Put $\alpha ={30}^{\circ }$, $\beta ={60}^{\circ }$ and $\theta ={90}^{\circ }$,

then a = $\frac{1}{2}$, b = $\frac{1}{2}$

∴ $co{s}^2(\alpha -\beta )+2absin(\alpha -\beta )$ = $\frac{3}{4}$ + $\frac{1}{2}$ × (- $\frac{1}{2}$) = $\frac{1}{2}$

which is given by option (c).