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Question: How do you find the value of \[{\sin ^2}\left( {{{225}^0}} \right)\] ?...

How do you find the value of sin⁡2(2250){\sin ^2}\left( {{{225}^0}} \right) ?

Explanation

Solution

Hint : Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions. We know that sin⁡2(θ){\sin ^2}\left( \theta \right) and (sin⁡(θ))2{\left( {\sin \left( \theta \right)} \right)^2} are the same. To solve this we need to know the sine supplementary angle.

Complete step-by-step answer :
We have,
sin⁡2(2250){\sin ^2}\left( {{{225}^0}} \right) .
First we find the value of sin⁡(2250)\sin \left( {{{225}^0}} \right) . Then we square the obtained answer.
We can write 225=180+45225 = 180 + 45 .
Then we have,
⇒sin⁡(2250)=sin⁡(1800+450)\Rightarrow \sin \left( {{{225}^0}} \right) = \sin \left( {{{180}^0} + {{45}^0}} \right)
We know that sin⁡(1800+θ)=−sin⁡θ\sin \left( {{{180}^0} + \theta } \right) = - \sin \theta , the negative sign is because the sine is negative in the third quadrant.
⇒sin⁡(2250)=sin⁡(1800+450)\Rightarrow \sin \left( {{{225}^0}} \right) = \sin \left( {{{180}^0} + {{45}^0}} \right)
=−sin⁡(450)= - \sin \left( {{{45}^0}} \right)
=−12= - \dfrac{1}{{\sqrt 2 }}
Thus we have,
⇒sin⁡(2250)=−12\Rightarrow \sin \left( {{{225}^0}} \right) = - \dfrac{1}{{\sqrt 2 }} .
Now squaring on both sides we have,
⇒(sin⁡(2250))2=(−12)2\Rightarrow {\left( {\sin \left( {{{225}^0}} \right)} \right)^2} = {\left( { - \dfrac{1}{{\sqrt 2 }}} \right)^2}
⇒sin⁡2(2250)=1(2)2\Rightarrow {\sin ^2}\left( {{{225}^0}} \right) = \dfrac{1}{{{{\left( {\sqrt 2 } \right)}^2}}}
We know that square and square root will cancels out,
⇒sin⁡2(2250)=12\Rightarrow {\sin ^2}\left( {{{225}^0}} \right) = \dfrac{1}{2} . This is the required answer.
So, the correct answer is “12\dfrac{1}{2}”.

Note : We can also solve this using the sum and difference formula of sine. That is sin⁡(a+b)=sin⁡(a).cos⁡(b)+cos⁡(a).sin⁡(b)\sin (a + b) = \sin (a).\cos (b) + \cos (a).\sin (b) and sin⁡(a−b)=sin⁡(a).cos⁡(b)−cos⁡(a).sin⁡(b)\sin (a - b) = \sin (a).\cos (b) - \cos (a).\sin (b) . We also have a cosine sum and the difference formula is cos⁡(a+b)=cos⁡(a).cos⁡(b)−sin⁡(a).sin⁡(b)\cos (a + b) = \cos (a).\cos (b) - \sin (a).\sin (b) and cos⁡(a−b)=cos⁡(a).cos⁡(b)+sin⁡(a).sin⁡(b)\cos (a - b) = \cos (a).\cos (b) + \sin (a).\sin (b) .
A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.