Question

if sin squared theta + 3 cos theta is equal to 2 then find the value of cos cube theta + sec cube theta

Answer 1

See the attachment for detail solution
Hope it will help you

Answer 2

Question: If $sin^2\theta +3cos\theta = 2$ then find the value of $cos^3\theta + sec^3\theta$.

Answer:

The value of $cos^3\theta + sec^3\theta$ is 18.

Step-by-step explanation:

It is given that $sin^2\theta +3cos\theta = 2$.

To find:-

The value of $cos^3\theta + sec^3\theta$.

Step 1 of 2

Consider the given trigonometric function as follows:

$sin^2\theta +3cos\theta = 2$ _____ (1)

As we know,

$sin^2\theta = 1-cos^2\theta$

Then, the equation (1) becomes,

$1-cos^2\theta +3cos\theta = 2$

$1-cos^2\theta +3cos\theta -2=0$

$cos^2\theta -3cos\theta +1=0$

Let $x$ = cos$\theta$. Then,

$x^2-3x+1=0$

Here, $a=1$, $b=-3$ and $c=1$

Notice that the equation is quadratic equation.

Solve the quadratic equation as follows:

$x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

$x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(1)} }{2(1)}$

  $=\frac{3\pm\sqrt{9-4} }{2}$

  $=\frac{3\pm\sqrt{5} }{2}$

$x=\frac{3+\sqrt{5} }{2}$ and $x=\frac{3-\sqrt{5} }{2}$

⇒ $cos\theta =\frac{3+\sqrt{5} }{2}$ and $cos\theta =\frac{3-\sqrt{5} }{2}$.

Case1. When $cos\theta =\frac{3+\sqrt{5} }{2}$. Then,

sec$\theta$ = 1/cos$\theta$

        = $1/\frac{3+\sqrt{5} }{2}$

        = $\frac{2}{3+\sqrt{5} }$

On rationalising, we get

sec$\theta$ = $\frac{2(3-\sqrt{5} )}{(3+\sqrt{5} )(3-\sqrt{5} )}$

        = $\frac{2(3-\sqrt{5}) }{4}$

sec$\theta$ = $\frac{3-\sqrt{5} }{2}$

Case2. When $cos\theta =\frac{3-\sqrt{5} }{2}$. Then,

sec$\theta$ = 1/cos$\theta$

        = $1/\frac{3-\sqrt{5} }{2}$

        = $\frac{2}{3-\sqrt{5} }$

On rationalising, we get

sec$\theta$ = $\frac{2(3+\sqrt{5} )}{(3+\sqrt{5} )(3-\sqrt{5} )}$

        = $\frac{2(3+\sqrt{5}) }{4}$

sec$\theta$ = $\frac{3+\sqrt{5} }{2}$

Step 2 of 2

Find the value of $cos^3\theta + sec^3\theta$.

Case1. When $cos\theta =\frac{3+\sqrt{5} }{2}$ and $sec\theta=\frac{3-\sqrt{5} }{2}$.

$cos^3\theta + sec^3\theta=\left(\frac{3+\sqrt{5} }{2} \right)^3+\left(\frac{3-\sqrt{5} }{2} \right)^3$

                     $=\frac{1}{8}[ \left(3+\sqrt{5} \right)^3+\left(3-\sqrt{5} \right)^3]$

                     $=\frac{1}{8}[ 27+5\sqrt{5} +9\sqrt{5}(3+\sqrt{5} ) +27-5\sqrt{5} -9\sqrt{5}(3-\sqrt{5} )]$

                     $=\frac{1}{8}[ 54 +90]$

                     $=\frac{1}{8}\times 144$

$cos^3\theta + sec^3\theta=18$

Case2. When $cos\theta =\frac{3-\sqrt{5} }{2}$ and $sec\theta=\frac{3+\sqrt{5} }{2}$.

$cos^3\theta + sec^3\theta=\left(\frac{3-\sqrt{5} }{2} \right)^3+\left(\frac{3+\sqrt{5} }{2} \right)^3$

                     $=\frac{1}{8}[ \left(3-\sqrt{5} \right)^3+\left(3+\sqrt{5} \right)^3]$

                     $=\frac{1}{8}[ 27-5\sqrt{5} -9\sqrt{5}(3-\sqrt{5} ) +27+5\sqrt{5} +9\sqrt{5}(3+\sqrt{5} )]$

                     $=\frac{1}{8}[ 54 +90]$

                     $=\frac{1}{8}\times 144$

$cos^3\theta + sec^3\theta=18$

Therefore, the value of $cos^3\theta + sec^3\theta$ is 18.

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