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Question: Solve \(\sin (2{\sin ^{ - 1}}0.6)\)...

Solve sin⁡(2sin⁡−10.6)\sin (2{\sin ^{ - 1}}0.6)

Explanation

Solution

First, we need to analyze the given information which is in the trigonometric form.
The trigonometric functions are useful whenever trigonometric values or functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
We also make use of the square of the numbers concept to solve further.
Formula used:
We know that the inverse of the sine formula for 2sin⁡−1x=sin⁡−1(2x1−x2)2{\sin ^{ - 1}}x = {\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} )

Complete step by step answer:
Since from the given that we have the sine value as sin⁡(2sin⁡−10.6)\sin (2{\sin ^{ - 1}}0.6).
Which can be converted into the form of 2sin⁡−1x=sin⁡−1(2x1−x2)2{\sin ^{ - 1}}x = {\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} )
Here the x=0.6x = 0.6 is the only difference in the given
Thus, we have the converted values as 2sin⁡−1(0.6)=sin⁡[sin⁡−1(2(0.6)1−(0.6)2)]2{\sin ^{ - 1}}(0.6) = \sin [{\sin ^{ - 1}}(2(0.6)\sqrt {1 - {{(0.6)}^2}} )]
Since we just need to know such things about the square root numbers and perfect square numbers and perfect cube numbers, A perfect square is the numbers that obtain by multiplying any whole numbers (zero to infinity) twice, or the square of the given numbers yields a whole number like 25=5\sqrt {25} = 5or 25=5225 = {5^2}
Hence, we have (0.6)2=0.36{(0.6)^2} = 0.36 and also using the multiplication operation we get 2×0.6=1.22 \times 0.6 = 1.2
Now substitute the values, we get sin⁡(2sin⁡−10.6)=sin⁡[sin⁡−1(1.2)1−0.36)]\sin (2{\sin ^{ - 1}}0.6) = \sin [{\sin ^{ - 1}}(1.2)\sqrt {1 - 0.36} )]
Since the inverse and the original values get cancel even in the trigonometry function, that is sin⁡(sin⁡−1)=1\sin ({\sin ^{ - 1}}) = 1 and hence we have sin⁡(2sin⁡−10.6)=(1.2)1−0.36\sin (2{\sin ^{ - 1}}0.6) = (1.2)\sqrt {1 - 0.36}
Further solving we have sin⁡(2sin⁡−10.6)=(1.2)0.64\sin (2{\sin ^{ - 1}}0.6)= (1.2)\sqrt {0.64}
Using the square root method, we get sin⁡(2sin⁡−10.6)=(1.2)×0.8\sin (2{\sin ^{ - 1}}0.6) = (1.2) \times 0.8
Hence, we get sin⁡(2sin⁡−10.6)=0.96\sin (2{\sin ^{ - 1}}0.6) = 0.96

Note:
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like sin⁡cos⁡=tan⁡\dfrac{{\sin }}{{\cos }} = \tan and tan⁡=1cot⁡\tan = \dfrac{1}{{\cot }}
We also used the multiplication operation, which is multiplier refers to multiplying the first number. Have a look at an example; while multiplying 5×75 \times 7the number 55 is called the multiplicand and the number 77 is called the multiplier. Like 2×3=62 \times 3 = 6 or which can be also expressed in the form of 2+2+2(3times)2 + 2 + 2(3times)