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Question: How do you evaluate \(\sin ( - {175^\circ }) \cdot \tan ({185^\circ }) \cdot \cos ({355^\circ }) + \...

How do you evaluate sin(175)tan(185)cos(355)+sin(85)cos(365)\sin ( - {175^\circ }) \cdot \tan ({185^\circ }) \cdot \cos ({355^\circ }) + \sin ( - {85^\circ }) \cdot \cos ({365^\circ }) ?

Explanation

Solution

In this question, we will use the basic identities of the trigonometric functions to find out the value of the given expression. We will rewrite the terms given in the expression as a sum of degrees by applying the properties of trigonometric function. Then we make use of periodicity of the trigonometric function and simplify the equation. We then substitute back the values to obtain the required solution.

Complete step-by-step solution:
Given an expression of the form sin(175)tan(185)cos(355)+sin(85)cos(365)\sin ( - {175^\circ }) \cdot \tan ({185^\circ }) \cdot \cos ({355^\circ }) + \sin ( - {85^\circ }) \cdot \cos ({365^\circ })
We are asked to evaluate the above expression.
We know that sin(A)=sinA\sin ( - A) = - \sin A
Hence the above expression can be written as,
sin(175)tan(185)cos(355)sin(85)cos(365)\Rightarrow - \sin ({175^\circ }) \cdot \tan ({185^\circ }) \cdot \cos ({355^\circ }) - \sin ({85^\circ }) \cdot \cos ({365^\circ })
Firstly, we will simplify the given expression by writing the trigonometric functions of the expression in the form of quadrants.
Note that we can write,
sin(175)=sin(1805)\sin ({175^\circ }) = \sin ({180^\circ } - {5^\circ })
tan(185)=tan(180+5)\tan ({185^\circ }) = \tan ({180^\circ } + {5^\circ })
cos(355)=cos(3605)\cos ({355^\circ }) = \cos ({360^\circ } - {5^\circ })
sin(85)=sin(905)\sin ({85^\circ }) = \sin ({90^\circ } - {5^\circ })
cos(365)=cos(360+5)\cos ({365^\circ }) = \cos ({360^\circ } + {5^\circ })
Therefore, we get,
sin(1805)tan(180+5)cos(3605)sin(905)cos(360+5)\Rightarrow - \sin ({180^\circ } - {5^\circ }) \cdot \tan ({180^\circ } + {5^\circ }) \cdot \cos ({360^\circ } - {5^\circ }) - \sin ({90^\circ } - {5^\circ }) \cdot \cos ({360^\circ } + {5^\circ })
Also we know that sin(180θ)=sinθ\sin ({180^\circ } - \theta ) = \sin \theta as in the second quadrant sin is positive, tan(180+θ)=tanθ\tan ({180^\circ } + \theta ) = \tan \theta as in the third quadrant tan is positive, cos(360θ)=cosθ\cos ({360^\circ } - \theta ) = \cos \theta as in the fourth quadrant cos is positive and cos(360+θ)=cosθ\cos ({360^\circ } + \theta ) = \cos \theta in the first quadrant.
Also we know that sin(90θ)=cosθ\sin ({90^\circ } - \theta ) = \cos \theta .
Using these results we get,
sin(5)tan(5)cos(5)cos(5)cos(5)\Rightarrow - \sin ({5^\circ }) \cdot \tan ({5^\circ }) \cdot \cos ({5^\circ }) - \cos ({5^\circ }) \cdot \cos ({5^\circ })
We know that tanθ=sinθcosθ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
Hence we get,
sin(5)sin(5)cos(5)cos(5)cos2(5)\Rightarrow - \sin ({5^\circ }) \cdot \dfrac{{\sin ({5^\circ })}}{{\cos ({5^\circ })}} \cdot \cos ({5^\circ }) - {\cos ^2}({5^\circ })
Simplifying we get,
sin(5)sin(5)cos2(5)\Rightarrow - \sin ({5^\circ }) \cdot \sin ({5^\circ }) - {\cos ^2}({5^\circ })
sin2(5)cos2(5)\Rightarrow - {\sin ^2}({5^\circ }) - {\cos ^2}({5^\circ })
Taking minus sign outside we get,
(sin2(5)+cos2(5))\Rightarrow - ({\sin ^2}({5^\circ }) + {\cos ^2}({5^\circ }))
We know the trigonometric identity, sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
Hence we get the final result as,
1\Rightarrow - 1
Hence the value of the expression sin(175)tan(185)cos(355)+sin(85)cos(365)\sin ( - {175^\circ }) \cdot \tan ({185^\circ }) \cdot \cos ({355^\circ }) + \sin ( - {85^\circ }) \cdot \cos ({365^\circ }) is 1 - 1.

Note: Students must know the basic properties of trigonometric functions and also in which quadrant which function is positive or negative.
As in the first quadrant all the six trigonometric functions are positive. In the second quadrant only the sine and cosec functions are positive, rest of all are negative. In the third quadrant, only the tan and cot functions are positive and all the other functions are negative. In the fourth quadrant only the cosine and secant are positive.