| If \[(sinx+\sin y)=a\]and \[(\cos x+\cos y)=b,\] what is the value of \[\sin x\sin y+\cos xcosy?\] |
A) \[a+b-ab\]
B) \[a+b+ab\]
C) \[{{a}^{2}}+{{b}^{2}}-2\]
D) \[\frac{{{a}^{2}}+{{b}^{2}}-2}{2}\]
E) None of these
Correct Answer: D
Solution :
| \[(\sin x+\sin y)=a\] |
| and \[(\cos x+\cos y)=b\] |
| On squaring both the equations, we get |
| \[{{(\sin x+\sin y)}^{2}}={{a}^{2}}\] |
| \[{{\sin }^{2}}x+{{\sin }^{2}}y+2\sin x\sin y={{a}^{2}}\] (i) |
| and \[{{(\cos x+\cos y)}^{2}}={{b}^{2}}\] |
| \[{{\cos }^{2}}x+{{\cos }^{2}}y+2\cos x\cos y={{b}^{2}}\] ... (ii) |
| On adding Eqs. (i) and (ii), we get |
| \[({{\sin }^{2}}x+{{\sin }^{2}}y+2\sin x\sin y)\] |
| \[+\,\,({{\cos }^{2}}x+{{\cos }^{2}}y+2\cos x\cos \,\,y)={{a}^{2}}+{{b}^{2}}\] |
| \[\Rightarrow \]\[{{\sin }^{2}}x+{{\cos }^{2}}x+{{\sin }^{2}}y+{{\cos }^{2}}y\] |
| \[+\,\,2\,\,(\sin x\sin y+\cos x\cos y)={{a}^{2}}+{{b}^{2}}\] |
| \[\Rightarrow \]\[1+1+2\,\,(\sin x\sin y+\cos x\cos y)={{a}^{2}}+{{b}^{2}}\] |
| \[\therefore \]\[\sin x\sin y+\cos x\cos y=\frac{{{a}^{2}}+{{b}^{2}}-2}{2}\] |
You need to login to perform this action.
You will be redirected in
3 sec
