Question

cos theta-4 sin theta=1 then sin theta +4 cos theta =?​

Answer 1

Given :

If cos $\sf\theta$ - 4 sin $\sf\theta$ = 1, then sin $\sf\theta$ + 4 cos $\sf\theta$ =

To Find :

To find the Value of sin $\sf\theta$ + 4 cos $\sf\theta$

Solution :

$\sf\ cos \: \theta \: - \: 4 \: sin \: \theta \: = \: 1$

$\sf\ ( \: cos \: \theta \: - \: 4 \: sin \: \theta \: )^{2} \: = \: ( \: 1 \: )^{2}$

${\large{\bf{⇒}}}\sf\ \: \: cos^{2} \: \theta \: + \: 16 \: sin^{2} \: \theta \: - \: 2 \: . \: cos \: \theta \: . \: 4 \: sin \: \theta \: = \: 1$

${\large{\bf{⇒}}}\sf\ \: \: cos {}^{2} \: \theta \: + \: {16 \: sin}^{2} \: \theta \: \: - \: 8 \: sin \: \theta \: . \: cos \: \theta \: = \: 1 \: \: \rule{25mm}{1pt} \: {\boxed{\sf{\red{1}}}}$

Now, Let sin $\theta$ + 4 cos $\theta$ = K

${\large{\bf{⇒}}}\sf\ \: ( \: sin \: \theta \: + \: 4 \: cos \: \theta \: ) {}^{2} \: = \: {k}^{2}$

${\large{\bf{⇒}}}\sf\ \: \: {sin}^{2} \: \theta \: + \: 16 \: {cos}^{2} \: \theta \: + \: 2 \: sin \: \theta \: . \: 4 \: cos \: \theta \: = \: {k}^{2}$

${\large{\bf{⇒}}}\sf\ \: \: {sin}^{2} \: \theta \: + \: 16 \: {cos}^{2} \: \theta \: + \: 8 \: sin \: \theta \: . \: cos \: \theta \: = \: {k}^{2} \: \: \rule{25mm}{1pt} \: {\boxed{\red{2}}}$

$\rule{100mm}{1pt}$

$\large\sf\ Now, \: Lets \: Add \: Eq \: {\boxed{\red{1}}} \: + \: {\boxed{\red{2}}}$

$\sf\ {cos}^{2} \: \theta \: + \: 16 \: {sin}^{2} \: \theta \: - \: \cancel{8 \: sin \: \theta \: cos \: \theta} \: = \: 1 \\ \frac{\sf\ {sin}^{2} \: \theta \: + \: 16 \: {cos}^{2} \: \theta \: + \: \cancel{8 \: sin \: \theta \: cos \: \theta} \: = \: {k}^{2} }{\sf\ ( \: {sin}^{2} \: \theta \: + \: {cos}^{2} \: \theta \: ) \: + \: 16 \: ( \: {sin}^{2} \: \theta \: + \: {cos}^{2} \: \theta \: ) \: = \: 1 \: + \: {k}^{2}}$

${\large{\bf{⇒}}}\sf\ \: \: 1 \: + \: 16 \: (1) \: = \: 1 \: + \: {k}^{2}$

${\large{\bf{⇒}}}\sf\ \: \: 18 \: = \: \cancel{1} \: + \: {k}^{2} \: \cancel{ - 1}$

${\large{\bf{⇒}}}\sf\ \: \: {k}^{2} \: = \: 16$

${\large{\bf{⇒}}}\sf\ \: \: k \: = \: \sqrt[±]{16}$

${\large{\bf{⇒}}}\sf\ \: \: k \: = \: ± \: 4.$

${\large{\boxed{\sf{\red{sin \: \theta \: + \: 4 \: cos \: \theta \: = \: ± \: 4.}}}}}$

$\rule{100mm}{1pt}$

Know More :

• The Meaning of '.' is '×'. (Multiplication)
• The Meaning of '±' is Plus or Minus.

Identidites :

$\sf\ \: \: sin^{2} \: \theta \: + \: cos {}^{2} \: \theta \: = \: 1$

$\sf\ \: \: sec {}^{2} \: \theta \: - \: tan {}^{2} \: \theta \: = \: 1$

$\sf\ \: \: {cosec}^{2} \: \theta \: - \: {cot}^{2} \: \theta \: = \: 1$

$\tiny$

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