Question

If $\sf{ Sin \; \theta }$ 1/2 than find the value of $\sf{ 3 \; Cos \; \theta }$ $\sf{ 4 \; {Cos}^{3} \; \theta }$ .​

Answer 1

Answer:

Solution

Solution3cosθ+4sinθ=5[35cosθ+45sinθ]

Solution3cosθ+4sinθ=5[35cosθ+45sinθ]=5[sinαcosθ+cosαsinθ]=5sin(θ+α)

Solution3cosθ+4sinθ=5[35cosθ+45sinθ]=5[sinαcosθ+cosαsinθ]=5sin(θ+α)Therefore A=5 and α=sin−1(35)

Step-by-step explanation:

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Answer 2

Given :

• $\sf{ Sin \; \theta = \dfrac{1}{2} }$

To Find :

• $\sf{ 3 \; Cos \; \theta - 4 \; {Cos}^{3} \; \theta }$

SolutioN :

In this question we need to find the Base by using Pythagoras theorem and after that we can derive the value . Let's Go :-

❍ Deriving the Value of θ :

• We know that :

$\; \; \dashrightarrow \; \; {\underline{\overline{\boxed{\red{\pmb{\sf{ Sin \; \theta = \dfrac{Perpendicular}{Height} }}}}}}}$

• Hence :

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { Sin \; \theta = \dfrac{Perpendicular}{Height} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\sf{ Sin \; \theta = \dfrac{1x}{2x} }}}} \; {\pink{\bigstar}} \\ \\ \\ \end{gathered}$

❍ Deriving the Base :

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { {Base}^{2} = {Hypotonous}^{2} - {Height}^{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { {Base}^{2} = {2x}^{2} - {1x}^{2} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { {Base}^{2} = 4x - 1x } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { {Base}^{2} = 3x } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\sf{ Base = \sqrt{3x} }}}} \; {\green{\bigstar}} \\ \\ \\ \end{gathered}$

❍ Deriving the Value of given Equation :

• We know That :

$\; \; \dashrightarrow \; \; {\underline{\overline{\boxed{\red{\pmb{\sf{ Cos \; \theta = \dfrac{Base}{Hypotonous} }}}}}}}$

• Hence :

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \sf{ 3 \; Cos \; \theta - 4 \; {Cos}^{3} \; \theta } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \sf{ 3 \times \dfrac{Base}{Hypotonous} - 4 \; { \dfrac{ Base }{ Hypotonous } }^{3} \; } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \sf{ 3 \times \dfrac{ \sqrt{3x} }{2x} - 4 \; { \dfrac{ \sqrt{3x} }{ 2x } }^{3} \; } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \sf{ 3 \times \dfrac{ \sqrt{3x} }{2x} - 4 \times \dfrac{3 \; \sqrt{3x} }{ 8x } \; } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \sf{ 3 \times \dfrac{ \sqrt{3x} }{2x} - \cancel4 \times \dfrac{3 \; \sqrt{3x} }{ \cancel{8x} } \; } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; \sf { \sf{ 3 \times \dfrac{ \sqrt{3x} }{2x} - \dfrac{3 \times \sqrt{3x} }{ 2x } \; } } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \; {\underline{\boxed{\sf{ 0 }}}} \; {\red{\bigstar}} \\ \\ \\ \end{gathered}$

• Therefore :

The Value of $\sf{ 3 \; Cos \; \theta - 4 \; {Cos}^{3} \; \theta }$ is 0 .

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