Question

If 7 cos theta - 24 sin theta = 0 then find cos theta + sin theta.​

Answer 1

GIVEN :–

$\\\bf \implies 7 \cos( \theta) - 24 \sin( \theta) = 0$

TO FIND :–

$\\\bf \implies \cos( \theta) + \sin( \theta) = ?$

SOLUTION :–

$\\\bf \implies 7 \cos( \theta) - 24 \sin( \theta) = 0$

• We know that –

$\\\bf \to \cos^{2} ( \theta) + \sin ^{2} ( \theta) = 1$

$\\\bf \to \cos^{2} ( \theta) = 1 -\sin ^{2} ( \theta)$

$\\\bf \to \cos( \theta) = \sqrt{1 -\sin ^{2} ( \theta)}$

• So that –

$\\\bf \implies 7\sqrt{1 -\sin ^{2} ( \theta)}- 24 \sin( \theta) = 0$

$\\\bf \implies 7\sqrt{1 -\sin ^{2} ( \theta)} = 24 \sin( \theta)$

• Square on both sides –

$\\\bf \implies 49(1 -\sin ^{2} \theta)= 576 \sin^{2} ( \theta)$

$\\\bf \implies 49 -49\sin ^{2} (\theta)= 576 \sin^{2} ( \theta)$

$\\\bf \implies 49 = 49\sin ^{2} (\theta) + 576 \sin^{2} ( \theta)$

$\\\bf \implies 49 =625 \sin^{2} ( \theta)$

$\\\bf \implies \sin^{2} ( \theta) = \dfrac{49}{625}$

$\\\bf \implies \sin( \theta) = \sqrt{ \dfrac{49}{625}}$

$\\\large \implies { \boxed{ \bf \sin( \theta) = \dfrac{7}{25}}}$

• And –

$\\\bf \implies \cos( \theta) = \sqrt{1 - \bigg( \dfrac{7}{25} \bigg)^{2}}$

$\\\bf \implies \cos( \theta) = \sqrt{1 - \bigg( \dfrac{49}{625} \bigg)}$

$\\\bf \implies \cos( \theta) = \sqrt{\dfrac{625 - 49}{625}}$

$\\\bf \implies \cos( \theta) = \sqrt{\dfrac{576}{625}}$

$\\\large \implies { \boxed{ \bf\cos( \theta) = \dfrac{24}{25}}}$

• Now –

$\\\bf \implies \cos( \theta) + \sin( \theta) = \dfrac{24}{25} + \dfrac{7}{25}$

$\\ \large \implies { \boxed{ \bf\cos( \theta) + \sin( \theta) = \dfrac{31}{25}}}$