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Question: How do you simplify \[\dfrac{{\sin \left( {\left( {\dfrac{\pi }{2}} \right) + h} \right) - \sin \lef...

How do you simplify sin⁡((π2)+h)−sin⁡(π2)h\dfrac{{\sin \left( {\left( {\dfrac{\pi }{2}} \right) + h} \right) - \sin \left( {\dfrac{\pi }{2}} \right)}}{h}?

Explanation

Solution

Here in this question, we have to simplify the given trigonometric form to the simplest form, this can be simplifying by using the sine addition identity in numerator i.e., sin⁡(a+b)=sin⁡a⋅cos⁡b+cos⁡a⋅sin⁡b\sin \left( {a + b} \right) = \sin a \cdot \cos b + \cos a \cdot \sin b on substituting and simplification we get the solution in the simplest form.

Complete step by step answer:
The question is related to trigonometry and it includes the trigonometry ratios. The trigonometry ratios are sine, cosine, tangent, cosecant, secant and cotangent. These trigonometry ratios are abbreviated as sin, cos, tan, csc, sec and cot.

Consider the given trigonometric function
sin⁡((π2)+h)−sin⁡(π2)h\dfrac{{\sin \left( {\left( {\dfrac{\pi }{2}} \right) + h} \right) - \sin \left( {\dfrac{\pi }{2}} \right)}}{h}-------(1)
Numerator having sin⁡((π2)+h)\sin \left( {\left( {\dfrac{\pi }{2}} \right) + h} \right) for finding this by using the sine addition identity i.e., sin⁡(a+b)=sin⁡a⋅cos⁡b+cos⁡a⋅sin⁡b\sin \left( {a + b} \right) = \sin a \cdot \cos b + \cos a \cdot \sin b
here, a=π2\dfrac{\pi }{2} and b=hh.
On using the sine addition identity the given inequality is written as
⇒  sin⁡((π2)+h)=sin⁡(π2)⋅cos⁡(h)+cos⁡(π2)⋅sin⁡(h)\Rightarrow \,\,\sin \left( {\left( {\dfrac{\pi }{2}} \right) + h} \right) = \sin \left( {\dfrac{\pi }{2}} \right) \cdot \cos \left( h \right) + \cos \left( {\dfrac{\pi }{2}} \right) \cdot \sin \left( h \right)
On substituting the value of sin⁡((π2)+h)\sin \left( {\left( {\dfrac{\pi }{2}} \right) + h} \right) in equation (1), we get
⇒   sin⁡(π2)⋅cos⁡(h)+cos⁡(π2)⋅sin⁡(h)−sin⁡(π2)h\Rightarrow \,\,\,\dfrac{{\sin \left( {\dfrac{\pi }{2}} \right) \cdot \cos \left( h \right) + \cos \left( {\dfrac{\pi }{2}} \right) \cdot \sin \left( h \right) - \sin \left( {\dfrac{\pi }{2}} \right)}}{h}-------(2)
By the standard trigonometric ratios angle table as we know the values of sin⁡(π2)=1\sin \left( {\dfrac{\pi }{2}} \right) = 1 and cos⁡(π2)=0\cos \left( {\dfrac{\pi }{2}} \right) = 0
On substituting these values in equation (2), we get
⇒   1⋅cos⁡(h)+0⋅sin⁡(h)−1h\Rightarrow \,\,\,\dfrac{{1 \cdot \cos \left( h \right) + 0 \cdot \sin \left( h \right) - 1}}{h}
On simplification, we get
⇒   cos⁡(h)−1h\Rightarrow \,\,\,\dfrac{{\cos \left( h \right) - 1}}{h}
We can’t simplify for the further and hence we obtained the result for the given question.

Hence, the simplest form of the given trigonometric function sin⁡((π2)+h)−sin⁡(π2)h\dfrac{{\sin \left( {\left( {\dfrac{\pi }{2}} \right) + h} \right) - \sin \left( {\dfrac{\pi }{2}} \right)}}{h} is cos⁡ h−1h\dfrac{{\cos \,h - 1}}{h}.

Note: Here the given question belongs to the topic trigonometry. In the question we have the word sin which means it is sine trigonometry ratio. Here we must know the trigonometric standard formula for the addition and subtraction. By the table of trigonometric ratios for the standard angles we simplify the given trigonometric function and hence we obtain the required result for the given question.