Question

Find the value of $\text{cos1755 }\!\!{}^\circ\!\!\text{ }$.

Answer 1

Hint: Cosine function is positive in first and fourth quadrant and negative in third and second quadrant. And if the angle inside the trigonometric function is type $\dfrac{n\pi }{2}\pm \theta $ (where n is an odd), then change cosine to sine, otherwise, if angle is 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 \[\text{cos}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=}\dfrac{\text{1}}{\sqrt{\text{2}}}\].

Complete step-by-step answer:

Here, we have to determine the value of the trigonometry term $\text{cos1755 }\!\!{}^\circ\!\!\text{ }$. So, let us suppose the value of the given trigonometric expression in the problem be A.

Hence, we can write equation as
$\text{A=1755 }\!\!{}^\circ\!\!\text{ }..............\text{(i)}$

Now, we can observe that the angle lying in $\text{0 }\!\!{}^\circ\!\!\text{ }\,\text{to}\,\,\text{90 }\!\!{}^\circ\!\!\text{ }$ i.e., not an acute angle and we have values of trigonometric functions only in $\text{0 }\!\!{}^\circ\!\!\text{ }\,\text{to}\,\,\text{90 }\!\!{}^\circ\!\!\text{ }$. 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: -

$1755{}^\circ =180{}^\circ \times 10{}^\circ -45{}^\circ .................(ii)$

Now as we know the radian representation of $180{}^\circ $ is given as
\[\pi \,\text{radian}=180{}^\circ ...........\text{(iii)}\]

Now, we can write $\text{45 }\!\!{}^\circ\!\!\text{ }$ by the above relation as-

$\begin{align}

  & {{180}^{{}^\circ }}=\pi \text{ radian} \\

 & \Rightarrow {{45}^{{}^\circ }}=\dfrac{\pi }{{{180}^{{}^\circ }}}\times {{45}^{{}^\circ }} \\

 & \Rightarrow {{45}^{{}^\circ }}=\dfrac{\pi }{4}\text{radian}...............(iv) \\

 & \\

\end{align}$

Now, we can put $1755{}^\circ $ as $10\pi -\dfrac{\pi }{4}\,$of the given expression in the problem i.e. in the equation (i). So, we can re-write the equation (i) as: -

\[A=\cos \left( 10\pi -\dfrac{\pi }{4} \right)...................(v)\]

Now, as we know the quadrant angles are defined as

Now, we can apply the trigonometric rules for the conversion of trigonometric expression by changing its angle.
Now, we can observe the equation (v) and get the angle involved with the trigonometric expression. i.e., $10\pi -\dfrac{\pi }{4}\,$ will lie in 4th quadrant.
Because $10\pi $is a multiple of $\text{2 }\!\!\pi\!\!\text{ }\,$and $\text{2 }\!\!\pi\!\!\text{ }\,$ lies at the same position where 0 lies, it means $10\pi $ will also lie there as well (is multiple of $\text{2 }\!\!\pi\!\!\text{ }\,$ and if we, rotate 5 times around the origin, then $10\pi $ will lie at same position where 0 lies\[\left( 10\pi =5\times 2\pi \right){}^\circ \] and we are subtracting $\dfrac{\pi }{4}$ from$10\pi $, it means the angle of $10\pi -\dfrac{\pi }{4}\,$ will lie in 4th quadrant as well. Now, as $10\pi -\dfrac{\pi }{4}\,$ is expressed in terms of angle $10\pi $, which is multiple of$\text{ }\!\!\pi\!\!\text{ }$; So, as per the trigonometric rules, the trigonometry function will not change with the expression which Is involved sum of angle which is multiple of$\text{ }\!\!\pi\!\!\text{ }$. It means, we can give some identities related to cos function and angle lying in 4th quadrant as
$\begin{align}
  & \cos \left( 2\pi +\theta \right)=\cos \theta \\
 & \cos \left( 4\pi +\theta \right)=\cos \theta \\

 & \cos \left( n\pi +\theta \right)=\cos \theta \\

 & . \\

 & . \\

 & . \\

 & \cos \left( n\pi -\theta \right)=\cos \theta \\

\end{align}$

Where, n is an even integer. So, with the help of above expression, we can write equation (v) as

$A=\cos \left( 10\pi -\dfrac{\pi }{4} \right)=\cos \dfrac{\pi }{4}$

Now, we know the value of $\cos \dfrac{\pi }{4}$ is$\dfrac{\text{1}}{\sqrt{\text{2}}}$. Hence, value of A is given as

$A=\dfrac{1}{\sqrt{2}}$

So, we get value of $\text{cos1755 }\!\!{}^\circ\!\!\text{ }$as

$\cos 1755{}^\circ =\dfrac{1}{\sqrt{2}}$

Hence, \[\dfrac{\text{1}}{\sqrt{\text{2}}}\]is the answer of the problem.

Note: One can express the trigonometric function $\cos 1755{}^\circ $ by following approach as $\cos 1755{}^\circ =\cos \left( 90{}^\circ \times 19{}^\circ +45{}^\circ \right)$

Be clear with the trigonometric rules. Take care of the conversion of trigonometric functions with the help of a sign of trigonometry functions in quadrants. Always try to write the angles higher than $\text{180 }\!\!{}^\circ\!\!\text{ }$ in terms of a sum of an acute and a multiple of$\text{180 }\!\!{}^\circ\!\!\text{ }$, so that it does not create any confusion because the trigonometric function will not change while converting the given angles of the form$n\pi \pm \theta $, here only needs to take care of sign. So, use the method applied for future reference as well.
Don’t go for calculating the exact value directly with any identity, it will be complex and may give the wrong answer as well. So, always try to solve these kinds of question by the way given in the solution.