Question

Find the value of $si{n}^2\theta +co{s}^2\theta at\theta ={30}^{\circ }\&{60}^{\circ }$?

• A. $\frac{1}{4},\frac{3}{4}$
• B. $\frac{3}{4},\frac{1}{4}$
• C. $1,1$
• D. $0,\frac{3}{4}$

Answer 1

The correct option is C

$1,1$

When $\theta ={30}^{\circ }$,

$sin\theta =\frac{1}{2}$ and $cos\theta =\frac{\sqrt{3}}{2}$,

$\therefore si{n}^2\theta +co{s}^2\theta$

$={\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2$

$=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1$

When $\theta ={60}^{\circ }$,

$sin\theta =\frac{\sqrt{3}}{2}$ and $cos\theta =\frac{1}{2}$,

$\therefore si{n}^2\theta +co{s}^2\theta$

$={\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2$

$=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1$

Or, Very simply, $si{n}^2\theta +co{s}^2\theta$ at all angles is = 1