Question

If $\tan \theta +\cot \theta =2$ where ${{0}^{\circ }}<\theta <{{90}^{\circ }}$, find the value of ${{\sin }^{15}}\theta +{{\cos }^{45}}\theta$.$$$$

Answer 1

We use the reciprocal relation between tangent and cotangent that is $\cot \theta =\dfrac{1}{\tan \theta }$ and p[proceed to simplify until we get a equation in the form whose $\tan \theta =\tan \alpha$ solutions are given by $x=n\pi +\alpha$ where $n$ is an integer. We find $\theta$ which satisfies ${{0}^{\circ }}< \theta <{{90}^{\circ }}$ and the put $\sin \theta ,\cos \theta$ in ${{\sin }^{15}}\theta +{{\cos }^{45}}\theta$ to obtain the required value.

Complete step-by-step answer:
We also know that the solutions of the equation $\tan x=\tan \alpha$ (where $x$ is the unknown variable and $\alpha$ is measure of angle) with arbitrary integer $n$are given by
$x=n\pi +\alpha$

We have the given equation from the question in tangent and cotangent as ,
$\tan \theta +\cot \theta =2$
We are also given the condition ${{0}^{\circ }} < \theta < {{90}^{\circ }}$ which means $\theta$ is an acute angle and so $\tan \theta ,\cot \theta$ are well defined here because $\tan \theta$does not exist for $\theta ={{90}^{\circ }}$ and $\cot \theta$ does not exist for $\theta ={{90}^{\circ }}$. We use the reciprocal relation between tangent and cotangent that is $\cot \theta =\dfrac{1}{\tan \theta }$ and proceed. We have
$\tan \theta +\dfrac{1}{\tan \theta }=2$
Let us multiply $\tan \theta$ both side of the equation and have,

& {{\tan }^{2}}\theta +1=2\tan \theta \\\ & \Rightarrow {{\tan }^{2}}\theta +1-2\tan \theta =0 \\\ \end{aligned}$$ Let us use the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$for $a=\tan \theta ,b=1$ and proceed to have, $$\Rightarrow {{\left( \tan \theta -1 \right)}^{2}}=0$$ We take square root both side of the above equation and get $$\begin{aligned} & \Rightarrow \tan \theta -1=0 \\\ & \Rightarrow \tan \theta =\tan \left( {{45}^{\circ }} \right) \\\ \end{aligned}$$ The solutions of the above equations are $$\begin{aligned} & \theta =n\pi +{{45}^{\circ }},n\in Z \\\ & \Rightarrow \theta =n{{180}^{\circ }}+{{45}^{\circ }} \\\ \end{aligned}$$ We are given the condition in the question that ${{0}^{\circ }} < \theta < {{90}^{\circ }}$. The only integer for which $\theta $ satisfies the given condition is $n=0$and hence $$\theta ={{45}^{\circ }}$$ We know that $$\sin \left( {{45}^{\circ }} \right)=\cos \left( {{45}^{\circ }} \right)=\dfrac{1}{\sqrt{2}}=\dfrac{1}{{{2}^{\dfrac{1}{2}}}}$$ We are asked to find the value of ${{\sin }^{15}}\theta +{{\cos }^{45}}\theta $ in the question. Let us put $\theta ={{45}^{\circ }}$ and have, $$\begin{aligned} & {{\sin }^{15}}{{45}^{\circ }}+{{\cos }^{45}}{{45}^{\circ }} \\\ & ={{\left( \sin {{45}^{\circ }} \right)}^{15}}+{{\left( \cos {{45}^{\circ }} \right)}^{45}} \\\ & ={{\left( \dfrac{1}{{{2}^{\dfrac{1}{2}}}} \right)}^{15}}+{{\left( \dfrac{1}{{{2}^{\dfrac{1}{2}}}} \right)}^{45}} \\\ \end{aligned}$$ We use the exponential identity ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ for $a=2,m=\dfrac{1}{2},n=15,45$ to have $$\begin{aligned} & =\dfrac{1}{{{2}^{\dfrac{1}{2}\times 15}}}+\dfrac{1}{{{2}^{\dfrac{1}{2}\times 45}}} \\\ & =\dfrac{1}{{{2}^{\dfrac{15}{2}}}}+\dfrac{1}{{{2}^{\dfrac{45}{2}}}} \\\ & =\dfrac{1}{{{2}^{7+\dfrac{1}{2}}}}+\dfrac{1}{{{2}^{22+\dfrac{1}{2}}}} \\\ \end{aligned}$$ We use the exponential identity ${{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}$to have $$\begin{aligned} & =\dfrac{1}{{{2}^{7+\dfrac{1}{2}}}}+\dfrac{1}{{{2}^{22+\dfrac{1}{2}}}} \\\ & =\dfrac{1}{{{2}^{7}}\cdot {{2}^{\dfrac{1}{2}}}}+\dfrac{1}{{{2}^{22}}\cdot {{2}^{\dfrac{1}{2}}}} \\\ \end{aligned}$$ We take ${{2}^{\dfrac{1}{2}}}$ common and then take ${{2}^{22}}$common and have, $$\begin{aligned} & =\dfrac{1}{{{2}^{\dfrac{1}{2}}}}\left( \dfrac{1}{{{2}^{7}}}+\dfrac{1}{{{2}^{22}}} \right) \\\ & =\dfrac{{{2}^{\dfrac{1}{2}}}}{2}\times \dfrac{1}{{{2}^{22}}}\left( {{2}^{15}}+1 \right) \\\ & =\dfrac{\sqrt{2}\left( {{2}^{15}}+1 \right)}{{{2}^{23}}} \\\ \end{aligned}$$ The above obtained value is the required value. $$$$ **Note:** The other relation between tangent and cotangent is complementary angle relation which is given by $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $. The solutions of $\tan x=0$ are integral multiples of $\pi $ and $\tan x$ does not exist for odd integral multiples of $\dfrac{\pi }{2}$.