Question

If a cos theta + b sin theta= c then a sin theta+ b cos theta= ? ​

Answer 1

Appropriate Question :-

$\rm :\longmapsto\:If \: acos\theta + bsin\theta = c \: then \: asin\theta - bcos\theta =$

$\large\underline{\sf{Solution-}}$

➢ Given that

$\rm :\longmapsto\:a \: cos\theta + b \: sin\theta = c$

On squaring both sides, we get

$\rm :\longmapsto\:(a \: cos\theta + b \: sin\theta)^{2} = c ^{2}$

We know,

$\underbrace{\boxed{\bf{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}}$

So, using this, we have

$\rm :\longmapsto\: {a}^{2} {cos}^{2}\theta + {b}^{2} {sin}^{2}\theta + 2absin\theta cos\theta = {c}^{2}$

We know,

$\underbrace{\boxed{\bf{ {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this

$\rm :\longmapsto\: {a}^{2}(1 - {sin}^{2}\theta) + {b}^{2}(1 - {cos}^{2}\theta )+ 2absin\theta cos\theta = {c}^{2}$

$\rm :\longmapsto\: {a}^{2}-{a}^{2} {sin}^{2}\theta+{b}^{2} - {b}^{2} {cos}^{2}\theta+ 2absin\theta cos\theta = {c}^{2}$

$\rm :\longmapsto\: -{a}^{2} {sin}^{2}\theta - {b}^{2} {cos}^{2}\theta+ 2absin\theta cos\theta ={c}^{2}-{a}^{2}-{b}^{2}$

$\rm :\longmapsto\: {a}^{2} {sin}^{2}\theta + {b}^{2} {cos}^{2}\theta - 2absin\theta cos\theta = - {c}^{2} + {a}^{2} + {b}^{2}$

We know,

$\underbrace{\boxed{\bf{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}}$

$\rm :\longmapsto\: {(asin\theta - bcos\theta )}^{2} = {a}^{2} + {b}^{2} - {c}^{2}$

$\bf :\longmapsto\:asin\theta - bcos\theta = \: \pm \: \sqrt{ {a}^{2} + {b}^{2} - {c}^{2} }$

Additional Information :-

$\underbrace{\boxed{\bf{ {sin}^{2}x + {cos}^{2}x = 1}}}$

$\underbrace{\boxed{\bf{ {sec}^{2}x - {tan}^{2}x = 1}}}$

$\underbrace{\boxed{\bf{ {cosec}^{2}x - {cot}^{2}x = 1}}}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$