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Question: Find the values of the trigonometric functions: i) \(\sin {{765}^{\circ }}\) ii) \(\csc (-1410)...

Find the values of the trigonometric functions:
i) sin⁡765∘\sin {{765}^{\circ }}
ii) csc⁡(−1410)\csc (-1410)

Explanation

Solution

Hint : Here, we know that the trigonometric functions like sine and cosec functions are periodic functions. Therefore, we can write sin⁡765∘\sin {{765}^{\circ }} as sin⁡765∘=sin⁡(2×360∘+45∘)\sin {{765}^{\circ }}=\sin (2\times {{360}^{\circ }}+{{45}^{\circ }}) and similarly we can write csc⁡(−1410)\csc (-1410) as, csc⁡(−1410)=−csc⁡(4×360∘−30∘)\csc (-1410)=-\csc (4\times {{360}^{\circ }}-{{30}^{\circ }})

Complete step by step solution :
Here, we have to find the value of sin⁡765∘\sin {{765}^{\circ }} and csc⁡(−1410)\csc (-1410).
i) sin⁡765∘\sin {{765}^{\circ }}
Now, first let us consider the function sin⁡765∘\sin {{765}^{\circ }}.
Here, first we should know about periodic functions.
We know that a periodic function is a function that repeats its value on regular intervals or periods. A function f is said to be periodic for a period t, if
f(x+t)=f(x)f(x+t)=f(x)
We also know that the trigonometric functions sin x, cos x and tan x are periodic functions. The functions sin x and cos x have the period 2π2\pi . Hence, we can say that,
sin⁡(2π+θ)=sin⁡θ cos⁡(2π+θ)=cos⁡θ \begin{aligned} & \sin (2\pi +\theta )=\sin \theta \\\ & \cos (2\pi +\theta )=\cos \theta \\\ \end{aligned}
Here, we are given sin⁡765∘\sin {{765}^{\circ }} and it can be written as:
sin⁡765∘=sin⁡(2×360∘+45∘)\sin {{765}^{\circ }}=\sin (2\times {{360}^{\circ }}+{{45}^{\circ }})
Since, sine function is periodic, we can say that,
sin⁡765∘=sin⁡45∘\sin {{765}^{\circ }}=\sin {{45}^{\circ }}
We know that the value of sin⁡45∘=12\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}.
Hence, we can say that, sin⁡765∘=12\sin {{765}^{\circ }}=\dfrac{1}{\sqrt{2}}

ii) csc⁡(−1410)\csc (-1410)
We also know that the cosec function, being a trigonometric function, is also a periodic function with period 2π2\pi . The function will repeat after the intervals 2π,4π,6π,..2\pi ,4\pi ,6\pi ,..
Here, we can write csc⁡(−1410)\csc (-1410) as:
csc⁡(−1410)=−csc⁡(1410)\csc (-1410)=-\csc (1410)
Since, we have csc⁡(−x)=−csc⁡x\csc (-x)=-\csc x.
We know that,
csc⁡(8π−x)=−csc⁡x ⇒csc⁡(4×2π−x)=−csc⁡x \begin{aligned} & \csc (8\pi -x)=-\csc x \\\ & \Rightarrow \csc (4\times 2\pi -x)=-\csc x \\\ \end{aligned}
Hence, we will get,
csc⁡(−1410)=−csc⁡(4×360∘−30∘)\csc (-1410)=-\csc (4\times {{360}^{\circ }}-{{30}^{\circ }})
Since, cosec function is a periodic function,
csc⁡(−1410)=−csc⁡(−30∘) ⇒csc⁡(−1410)=csc⁡(30∘) \begin{aligned} & \csc (-1410)=-\csc (-{{30}^{\circ }}) \\\ & \Rightarrow \csc (-1410)=\csc ({{30}^{\circ }}) \\\ \end{aligned}
We have,
csc⁡30∘=2\csc {{30}^{\circ }}=2
Therefore, we will get the function as,
⇒csc⁡(−1410)=2\Rightarrow \csc (-1410)=2
Hence, we can say that the value of csc⁡(−1410)=2\csc (-1410)=2.

Note : Students generally get confused in the trigonometric formulae. Students get confused and write sin⁡(2π+θ)=cos⁡θ\sin (2\pi +\theta )=\cos \theta and cos⁡(2π+θ)=sin⁡θ\cos (2\pi +\theta )=\sin \theta , which is wrong. This confusion should be avoided as it can lead to wrong answers. The formulae must be remembered properly.