Given $\cos (A - B) = \cos A\cos B + \sin A\sin B$ . Taking suitable A and B, find $\cos {15^ \circ }$ .
Answer
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Hint: Given in the equation is a linear equation with the trigonometric terms in it. If the above example is considered, A and B are the variables whose difference is 15. In all such sums of trigonometry, where the angle is unknown always take help of the standard angles which are used in Trigonometry. We just have to find the suitable values of A and B and substitute in the equation given.
Complete answer: Equation given in the question:
$\cos (A - B) = \cos A\cos B + \sin A\sin B$ ---1
To find: $\cos {15^ \circ }$
We need to find $\cos {15^ \circ }$ , so we start with the LHS of the above equation, i.e. $\cos (A - B)$
We need to understand here that A and B are the angles. So, according to the trigonometric concepts, angles can possibly be ${90^ \circ },{60^ \circ },{30^ \circ },{45^ \circ },{0^ \circ }$
To find, $\cos {15^ \circ }$ we need to have A and B in such a manner that \[A - B = 15\]
Therefore, let \[A = 60{\text{ }}and{\text{ }}B = 45\]
Substituting values of A and B in equation 1, we get,
$\cos ({60^ \circ } - {45^ \circ }) = \cos {60^ \circ }\cos {45^ \circ } + \sin {60^ \circ }\sin {45^ \circ }$
RHS: $\cos {60^ \circ }\cos {45^ \circ } + \sin {60^ \circ }\sin {45^ \circ }$
Putting the values of these trigonometric functions, we get,
$\dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} + \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} = \dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }}$
Equating LHS to RHS, we get,
$\cos {15^ \circ } = \dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }}$
Therefore, $\cos {15^ \circ } = \dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }}$.
Note:
Few things should be kept in mind when we come across such questions. Firstly, the concepts of trigonometric equations must be well known. In the above example, we have selected the values of A and B on the basis of their difference. From the angles, ${90^ \circ },{60^ \circ },{30^ \circ },{45^ \circ },{0^ \circ }$ , difference of 15 can be obtained by 2 combinations, 60-45 and 45-30. The above sum is solved using 60-45. You can also try using A=45 and B=30. Both will give the same answer.
Complete answer: Equation given in the question:
$\cos (A - B) = \cos A\cos B + \sin A\sin B$ ---1
To find: $\cos {15^ \circ }$
We need to find $\cos {15^ \circ }$ , so we start with the LHS of the above equation, i.e. $\cos (A - B)$
We need to understand here that A and B are the angles. So, according to the trigonometric concepts, angles can possibly be ${90^ \circ },{60^ \circ },{30^ \circ },{45^ \circ },{0^ \circ }$
To find, $\cos {15^ \circ }$ we need to have A and B in such a manner that \[A - B = 15\]
Therefore, let \[A = 60{\text{ }}and{\text{ }}B = 45\]
Substituting values of A and B in equation 1, we get,
$\cos ({60^ \circ } - {45^ \circ }) = \cos {60^ \circ }\cos {45^ \circ } + \sin {60^ \circ }\sin {45^ \circ }$
RHS: $\cos {60^ \circ }\cos {45^ \circ } + \sin {60^ \circ }\sin {45^ \circ }$
Putting the values of these trigonometric functions, we get,
$\dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }} + \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} = \dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }}$
Equating LHS to RHS, we get,
$\cos {15^ \circ } = \dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }}$
Therefore, $\cos {15^ \circ } = \dfrac{{1 + \sqrt 3 }}{{2\sqrt 2 }}$.
Note:
Few things should be kept in mind when we come across such questions. Firstly, the concepts of trigonometric equations must be well known. In the above example, we have selected the values of A and B on the basis of their difference. From the angles, ${90^ \circ },{60^ \circ },{30^ \circ },{45^ \circ },{0^ \circ }$ , difference of 15 can be obtained by 2 combinations, 60-45 and 45-30. The above sum is solved using 60-45. You can also try using A=45 and B=30. Both will give the same answer.
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