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Question: If \[4{\cos ^2}A - 3 = 0\] and \[{0^ \circ } \leqslant A \leqslant {90^ \circ }\], then find \[\cos ...

If 4cos2A3=04{\cos ^2}A - 3 = 0 and 0A90{0^ \circ } \leqslant A \leqslant {90^ \circ }, then find cos3A\cos 3A.

Explanation

Solution

In order to determine the trigonometric function, cos3A\cos 3A. Here, the given function is 4cos2A3=04{\cos ^2}A - 3 = 0 and the degrees are 0A90{0^ \circ } \leqslant A \leqslant {90^ \circ }. First we need to know the trigonometric degree value, cos30=32\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}.

Complete step by step answer:
Given,
4cos2A3=04{\cos ^2}A - 3 = 0
And the another given condition is 0A90{0^ \circ } \leqslant A \leqslant {90^ \circ },
That is, it remains in the first quadrant where all the values are to be positive, no negative values occur.
Hence the following step is,
4cos2A3=04{\cos ^2}A - 3 = 0
Bringing 3 to the other side of the equation,
4cos2A=34{\cos ^2}A = 3
Equating the equation as making the term cos2A{\cos ^2}Ato be on the one side,
cos2A=34{\cos ^2}A = \dfrac{3}{4}
To find a single value, taking square root on both the sides,
cosA=34\cos A = \sqrt {\dfrac{3}{4}}
As the square root taken, the value becomes
cosA=±32\cos A = \pm \dfrac{{\sqrt 3 }}{2}
Neglecting the above negative values, in order of first quadrant where the angle present only in the 0A90{0^ \circ } \leqslant A \leqslant {90^ \circ }, hence the value becomes
cosA=32\cos A = \dfrac{{\sqrt 3 }}{2}
Based on the trigonometric values in the cosine angle, cos30=32\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}
Thus,
cosA=cos30\cos A = \cos {30^ \circ }
Equating on both the sides, the value of A is,
A=30A = {30^ \circ }
And we need to find the value of cos3A\cos 3A

cos3A=cos3(30) cos3A=cos90  \cos 3A = \cos 3({30^ \circ }) \\\ \cos 3A = \cos {90^ \circ } \\\

Hence multiplying the above value with the three times of its cosine value we found the cos3A\cos 3A.
Therefore,
cos3A=cos90\cos 3A = \cos {90^ \circ }
From the above question of the trigonometric values, the value of the given question given with some conditions which are as,
4cos2A3=04{\cos ^2}A - 3 = 0
0A90{0^ \circ } \leqslant A \leqslant {90^ \circ }
Which indicates the angle of the value lies in the first quadrant and need to find the value of cos3A\cos 3A
Hence the solution is, cos3A=cos90\cos 3A = \cos {90^ \circ }

Note: We note that the given problem needs to convert the values into cosine degrees from the degree table. We know that the trigonometric degree table.
There are eight Trigonometric identities called fundamental identities. Three of them are called Pythagorean identities as they are based on Pythagorean Theorem.