Question
If A = B = 60°, verify that.
$\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}$
$\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}$
✓
Answer
Given:
$\text{A}=\text{B}=60^\circ\ \dots(1)$
To verify:
$\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}\ \dots(2)$
Now consider left hand side of the expression to be verified in equation (2)
Therefore,
$\cos(\text{A}-\text{B})=\cos(60-60)$
$=\cos0$
$=1$
Now consider right hand side of the expression to be verified in equation (2)
Therefore,
$\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}=\cos\text{B}\cos\text{B}+\sin\text{B}\sin\text{B}$
$=\cos^2\text{B}+\sin^2\text{B}$
$=1$
Hence it is verified that,
$\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}$
$\text{A}=\text{B}=60^\circ\ \dots(1)$
To verify:
$\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}\ \dots(2)$
Now consider left hand side of the expression to be verified in equation (2)
Therefore,
$\cos(\text{A}-\text{B})=\cos(60-60)$
$=\cos0$
$=1$
Now consider right hand side of the expression to be verified in equation (2)
Therefore,
$\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}=\cos\text{B}\cos\text{B}+\sin\text{B}\sin\text{B}$
$=\cos^2\text{B}+\sin^2\text{B}$
$=1$
Hence it is verified that,
$\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}$
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