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Question: Find the values of i) \(\sin 15^\circ \) ii) \(\cos 75^\circ \) iii) \(\tan 105^\circ \) iv) \(...

Find the values of
i) sin⁡15∘\sin 15^\circ ii) cos⁡75∘\cos 75^\circ iii) tan⁡105∘\tan 105^\circ
iv) cot⁡225∘\cot 225^\circ

Explanation

Solution

Hint: To solve this question we will use the identities of sin⁡(A+B)\sin (A + B) and cos⁡(A+B)\cos (A + B). Also, we will use the value of trigonometric functions at 30∘30^\circ , 45∘45^\circ , etc after converting the given angles in the known angles.

Complete step-by-step answer:
Now, we will first find the value of sin⁡15∘\sin 15^\circ . Now, sin⁡15∘\sin 15^\circ can be written as sin⁡15∘=sin⁡(45∘−30∘)\sin 15^\circ = \sin (45^\circ - 30^\circ )
Now, we will use the identity of sin⁡(A+B)\sin (A + B). From trigonometry, we have
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B
Similarly, from the above identity, the value of sin⁡(A−B)\sin (A - B) can be written as,
sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A\cos B - \cos A\sin B
So, sin⁡15∘=sin⁡(45∘−30∘)\sin 15^\circ = \sin (45^\circ - 30^\circ ) = sin⁡45∘cos⁡30∘−cos⁡45∘sin⁡30∘\sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ
Applying values of trigonometric functions at 30∘30^\circ and 45∘45^\circ , we get
sin⁡15∘=12.32−12.12\sin 15^\circ = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2}
So, sin⁡15∘=3−122\sin 15^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}
Now, we will use the identity cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B to find the value of cos⁡75∘\cos 75^\circ .
Now, cos⁡75∘\cos 75^\circ = cos⁡(45∘+30∘)\cos (45^\circ + 30^\circ ) = cos⁡45∘cos⁡30∘−sin⁡45∘sin⁡30∘\cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ
cos⁡75∘=12.32−12.12\cos 75^\circ = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2}
Therefore, cos⁡75∘=3−122\cos 75^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}
Now, tan⁡(A+B)=sin⁡(A+B)cos⁡(A+B)\tan (A + B) = \dfrac{{\sin (A + B)}}{{\cos (A + B)}}
So, tan⁡105∘=tan⁡(60∘+45∘)=sin⁡(60∘+45∘)cos⁡(60∘+45∘)\tan 105^\circ = \tan (60^\circ + 45^\circ ) = \dfrac{{\sin (60^\circ + 45^\circ )}}{{\cos (60^\circ + 45^\circ )}}
tan⁡105∘=32.12+12.1212.12−32.12\tan 105^\circ = \dfrac{{\dfrac{{\sqrt 3 }}{2}.\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2}.\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{2}.\dfrac{1}{{\sqrt 2 }} - \dfrac{{\sqrt 3 }}{2}.\dfrac{1}{{\sqrt 2 }}}} = 3+11−3\dfrac{{\sqrt 3 + 1}}{{1 - \sqrt 3 }}
tan⁡105∘=−(3+13−1)\tan 105^\circ = - (\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}})
Now, cot⁡225∘\cot 225^\circ can be written as cot⁡(180∘+45∘)\cot (180^\circ + 45^\circ ), so it lies in the third quadrant . So, we get
cot⁡225∘=cot⁡45∘\cot 225^\circ = \cot 45^\circ
Now, we will use the value of cot x at 45∘45^\circ . As, cot⁡45∘=1\cot 45^\circ = 1
Therefore, cot⁡225∘=1\cot 225^\circ = 1
So, sin⁡15∘=3−122\sin 15^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}
cos⁡75∘=3−122\cos 75^\circ = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}
tan⁡105∘=−(3+13−1)\tan 105^\circ = - (\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}})
cot⁡225∘=1\cot 225^\circ = 1

Note: Whenever we come up with such types of questions, we will use the properties of trigonometry. Such questions can be easily solved when we remember the term Add Coffee To Sugar which represents all, sin, tan and cos. This term shows in which quadrant which function remains positive. For example, in the first quadrant, all functions are positive, in the second quadrant, only sin and cosec are positive and so on. This is used in almost every problem and helps in solving difficult problems easily. Also, we have used the identities of sin⁡(A+B)\sin (A + B) and cos⁡(A+B)\cos (A + B)in this question. These identities are basic identities and all other identities can be derived from them.