Question

Find the value of:
(i) \[\sin {75^ \circ }\]
(ii) \[\tan {15^ \circ }\]

Answer 1

Hint: According to the question, here, in \[\sin {75^ \circ }\] divide 75 into two parts of which the values are known that is \[\sin {(45 + 30)^ \circ }\]. Similar in \[\tan {15^ \circ }\]that is \[\tan {15^ \circ } = \tan ({45^ \circ } - {30^ \circ })\].Hence, use the trigonometric formulas to solve it.

Formula Used:
 Here, we can use the trigonometric formulas that are:
1. \[\sin (A + B) = \sin A\cos B + \cos A\sin B\]
2. \[\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]

Complete step-by-step answer:
(i) \[\sin {75^ \circ } = \sin {(45 + 30)^ \circ }\]
Here, we will use the formula that is \[\sin (A + B) = \sin A\cos B + \cos A\sin B\] where \[A = {45^o}\] and \[B = {30^o}\]
On substituting the values of A and B we get,
\[\sin {(45 + 30)^ \circ } = \sin {45^ \circ }\cos {30^ \circ } + \cos {45^ \circ }\sin {30^ \circ }\]
As we know, \[\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\] , \[\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}\] , \[\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\] and \[\sin {30^ \circ } = \dfrac{1}{2}\]
Now, putting all the values in above equation:
\[ \Rightarrow \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}\]
Multiplying all the values where given,
\[ \Rightarrow \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }}\]
By taking L.C.M we get,
\[ \Rightarrow \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}\]
(ii) \[\tan {15^ \circ } = \tan ({45^ \circ } - {30^ \circ })\]
Here, we will use the formula that is \[\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\] where \[A = {45^o}\] and \[B = {30^o}\]
On substituting the values of A and B we get,
\[\tan ({45^ \circ } - {30^ \circ }) = \dfrac{{\tan {{45}^ \circ } - \tan {{30}^ \circ }}}{{1 + \tan {{45}^ \circ }\tan {{30}^ \circ }}}\]
As we know, \[\tan {45^ \circ } = 1\] and \[\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}\]
Now, putting all the values in above equation:
\[ \Rightarrow \dfrac{{1 - \dfrac{1}{{\sqrt 3 }}}}{{1 + 1 \times \dfrac{1}{{\sqrt 3 }}}}\]
On Multiplying and by taking L.C.M we get,
\[ \Rightarrow \dfrac{{\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 }}}}\]
Cancelling \[\sqrt 3 \] from both numerator and denominator we get,
\[ \Rightarrow \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\]
Hence, \[\sin {75^ \circ } = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}\] and \[\tan {15^ \circ } = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\]

Additional information:
There can be similar types of questions of trigonometry in which trigonometric values are different, that is instead of sin there can be cos or tan and vice-versa. To solve it, we use trigonometric formulas in which A and B occurs. Easy way to solve the trigonometric questions is to just convert them into cos and sin respectively.

Note: To solve these types of questions, firstly check which trigonometric formula is applicable and put the respective values of trigonometric ratios correctly. Then simplify the equations to get the required values. Hence, the trigonometric formulas make the question simpler and easy to understand.