Question

Find the value of Trigonometric expression $\cos \left( -1125{}^\circ \right)$ .

Answer 1

Hint: Use the trigonometric identity $\cos \left( -\theta \right)=\cos \theta$ to get a familiar relation. Now, cos function is positive in 1st and (iv)th quadrant and negative in 2nd and 3rd quadrant. And if the angle inside the trigonometric function is of type $\dfrac{n\pi }{2}\pm \theta$ (where n is an odd), then change cos to sin, otherwise, if angle of type $n\pi \pm \theta$ , then do not change the trigonometric function, use these rules to solve the given problem. Use the value of $\cos \left( \dfrac{\pi }{4} \right)=\left( \dfrac{1}{\sqrt{2}} \right)$.

Complete step-by-step answer:
Here, we have to determine the value of the trigonometric term $\cos \left( -1125{}^\circ \right)$.
So, let us suppose the value of the given trigonometric expression in the problem be ‘A’.
Hence, we can write equation as –
$A=\cos \left( -1125{}^\circ \right)$ …………………….. (i)
Now, as the angle inside the expression is negative. So, we need to use the following trigonometric identity of cosine functions, given as –
$\cos \left( -\theta \right)=\cos \theta$……………………… (ii)
So, using the above expression, we can re-write the equation (i) as –
$A=\cos \left( -1125{}^\circ \right)=\cos 1125{}^\circ$
Or $A=\cos 1125{}^\circ$ ……………………….. (iii)
Now, we can observe that the angle involved in the above expression is not lying in $0{}^\circ$ to ${{90}^{\circ }}$. It means we have to convert the given angle to acute angle form with the help of some trigonometric identities.
Hence, Let us divide the given expression by ${{180}^{\circ }}$ . So, that we can write the given angle in form of sum of angle which is multiple of ${{180}^{\circ }}$in following way:-
So, let us divide ${{1125}^{\circ }}$ by ${{180}^{\circ }}$ as,

& 1125 \\\ & \underline{1080} \\\ & \underline{ 45} \\\ \end{aligned}}\right.}}$$ So, we can write the angle ${{1125}^{\circ }}$ as ${{1125}^{\circ }}={{180}^{\circ }}\times 6+45{}^\circ $ ………………………….(iv) Now, as we know the radian representation of $180{}^\circ $ is given as – $\pi $ radian $=180{}^\circ $ …………………….. (v) And with the help of above relation, we can re-write the angle ${{45}^{\circ }}$ in radian form as well, given as – ${{45}^{\circ }}=\dfrac{\pi }{4}radian$ …………… (vi) Hence, we can re-write the expression (iv) with the help of equations (v) and (vi). ${{1125}^{\circ }}=6\pi +\dfrac{\pi }{4}$ …………………………. (vii) Now, we can put the angle ${{1125}^{\circ }}$ as $6\pi +\dfrac{\pi }{4}$ to the equation (iii). So, we get – $A=\cos \left( 6\pi +\dfrac{\pi }{4} \right)$ ……………….. (viii) Now, as we know the quadrant angles are defined as  Now, we can apply the trigonometric rules for conversion of trigonometric expressions by changing it’s angle. We can observe the equation (viii) and the given quadrants and hence, get that the angle $6\pi $ will lie at the same position where ${{0}^{\circ }}$ or $2\pi $is lying in the third rotation of ${{360}^{\circ }}$ or $2\pi $angle around the quadrant $\left( 6\pi =2\pi \times 3 \right)$ . Hence, the angle $\left( 6\pi +\dfrac{\pi }{4} \right)$ will lie in the 1st quadrant. As we know all the trigonometric functions are positive in the first quadrant and as the angle $\left( 6\pi +\dfrac{\pi }{4} \right)$ involved angle $6\pi $ which is multiple of $\pi $ , so trigonometric function will not change as per the rules of conversion of trigonometric functions. So, we can give relation as – $\begin{aligned} & \cos \left( 2\pi +\theta \right)=\cos \theta \\\ & \cos \left( 4\pi +\theta \right)=\cos \theta \\\ & \cos \left( 6\pi +\theta \right)=\cos \theta \\\ \end{aligned}$ Or $\cos \left( n\pi +\theta \right)=\cos \theta $………………………………. (ix) Where n is an even integer. Hence, we get the equation (viii) from the result (ix) as – $A=\cos \left( 6\pi +\dfrac{\pi }{4} \right)=\cos \dfrac{\pi }{4}$ ……………………………. (x) Now, we know the value of cos$\dfrac{\pi }{4}$ or $\cos {{45}^{\circ }}$ is given as $\dfrac{1}{\sqrt{2}}$ . Hence, the equation (x) can be given as $A=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ . So, $\dfrac{1}{\sqrt{2}}$is the answer to the problem. Note: One may get confuse the identity $\cos \left( -\theta \right)=\cos \theta $ by the other relations for other trigonometric functions. For future reference, other trigonometric relations are given as – $\begin{aligned} & \sin \left( -x \right)=-\sin x,\tan \left( -x \right)=-\tan x \\\ & \cos ec\left( -x \right)=-\cos ecx,\sec \left( -x \right)=-\sec x \\\ & \cot \left( -x \right)=-\cot x \\\ \end{aligned}$ One may not be able to find the direct value of $\cos \left( -1125{}^\circ \right)$ or not be able to convert $-1125{}^\circ $as done with $1125{}^\circ $ . So, we need to use $\cos \left( -\theta \right)=\cos \theta $ to get the familiar form of trigonometric function. So, don’t confuse with the negative sign inside the trigonometric function.