Question

How do you simplify \[\cos \left( t-\dfrac{\pi }{2} \right)\] ?

Answer 1

Hint: In this question we are asked to find the solution of \[\cos \left( t-\dfrac{\pi }{2} \right)\]. So, for solving this question we will use the basic concept of trigonometry and its formula that is \[\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b\].
So, we will solve the question by assuming a as t and b as \[\dfrac{\pi }{2}\] and substitute in the above formula and find its solution.

Complete step by step solution:
Firstly, the given question is in form of \[\cos \left( a-b \right)\] so by using its formula \[\cos \left( a-b \right)=\cos a\cos b+\sin a\sin b\] we will solve the question as follows.
Let us assume a as t and b as \[\dfrac{\pi }{2}\] and substitute in the above formula. So, the equation will be reduced as follows.
\[\Rightarrow \cos \left( a-b \right)=\cos a\cos b+\sin a\sin b\]
\[\Rightarrow \cos \left( t-\dfrac{\pi }{2} \right)=\cos t\cos \dfrac{\pi }{2}+\sin t\sin \dfrac{\pi }{2}\]
Here we know that the value of \[\cos \dfrac{\pi }{2}\] and \[\sin \dfrac{\pi }{2}\] respectively are 0 and 1. So substituting the values in the above equation the equation will be reduced as follows.
\[\Rightarrow \cos \left( t-\dfrac{\pi }{2} \right)=\cos t\times 0+\sin t\times 1\]
Here after substituting the values we simplify or do the calculation to the above equation. So, after doing the calculation the equation will be reduced as follows.
\[\Rightarrow \cos \left( t-\dfrac{\pi }{2} \right)=\sin t\]

So, the solution to the above question will be \[\sin t\].

Note: Students must perform the calculations very carefully. Students must be having good knowledge in the concept of trigonometry and its application. We must also have good grip in the formulae of trigonometry.
We can also solve this question without using the trigonometry formula by using the concept of quadrants in the trigonometry. Another solution to this question rather than using formula will be as follows.
\[ \cos \left( t-\dfrac{\pi }{2} \right)\] this expression looks alike as \[ \cos \left( \theta -\dfrac{\pi }{2} \right)\] so this lies in the first quadrant. We know that \[ \cos \left( \theta -\dfrac{\pi }{2} \right)\] will be \[\sin \left( \theta \right)\]. But \[ \cos \left( \theta -\dfrac{\pi }{2} \right)\] lies in first quadrant, So the solution \[\sin \left( \theta \right)\] will be positive which is +\[\sin \left( \theta \right)\]