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Select problems from miscellaneous exercise

Solutions to few problems from miscellaneous exercise on class 11 trigo.
In this article we will look at solutions of a few selected problems from miscellaneous exercise on chapter 3 of NCERT.
Problem 1:
Prove that:
2cosπ13cos9π13+cos3π13+cos5π13=0
Solution:
See the expression on the LHS. How do we start simplifying things? We have a couple of options.
Breakdown 2cosπ13cos9π13 using the formula
2cosAcosB=cos(A+B)+cos(AB)
Or, evaluate cos3π13+cos5π13 using the formula
cosA+cosB=2cosA+B2cosAB2
Let's try the second option.
=cos3π13+cos5π13=2cos4π13cos(π13)=2cos4π13cosπ13
Now, the LHS becomes
=2cosπ13cos9π13+2cos4π13cosπ13=2cosπ13(cos9π13+cos4π13)
Now recall that cos(πx)=cosx. Therefore,
cos9π13=cos(π4π13)=cos4π13
Using this in the expression above, we have
=2cosπ13(cos9π13+cos4π13)=2cosπ13(cos4π13+cos4π13)=0=RHS
Try to prove this again by proceeding via the first option.
Problem 2:
Prove that:
(cosx+cosy)2+(sinxsiny)2=4cos2x+y2
Solution:
Let's expand the LHS.
=(cosx+cosy)2+(sinxsiny)2=cos2x+cos2y+2cosxcosy+sin2x+sin2y2sinxsiny=sin2x+cos2x+sin2y+cos2y+2(cosxcosysinxsiny)=1+1+2cos(x+y)=2[1+cos(x+y)]=4cos2x+y2cos(x+y)=2cos2(x+y2)1=RHS
Problem 3:
Prove that:
sinx+sin3x+sin5x+sin7x=4cosxcos2xsin4x
Solution:
How do we make LHS equal to RHS? The most obvious thing which comes to mind is to apply the formula sinA+sinB=2sinA+B2cosAB2 on the LHS.
Let's do that.
=sinx+sin3x+sin5x+sin7x=2sin2xcosx+2sin6xcosx=2cosx(sin2x+sin6x)=2cosx(2sin4xcos2x)=4cosxcos2xsin4x=RHS
Problem 4:
Find sinx2, cosx2 and tanx2 for the following case:
tanx=43, x in quadrant II
Solution:
First let's try to find cosx. Why? If we know cosx, we can easily get cosx2 and sinx2 by using the identity
cosx=cos2x2sin2x2=2cos2x21=12sin2x2
After that we can easily find tanx2 by using tanx2=sinx2cosx2.
In given question, x lies in quadrant II.
π2xππ4x2π2
So, x2 will lie in quadrant I.
Now
sec2x=1+tan2x=1+169=259cos2x=925cosx=35 or35
Because x lies in quadrant II, cosx is negative. So, cosx=35.
2cos2x21=35cos2x2=15cosx2=15 or15
Because x2 lies in quadrant I, cosx2 is positive. So, cosx2=15. Similarly,
12sin2x2=35sin2x2=45sinx2=25 or25
Because x2 lies in quadrant I, sinx2 is positive. So, sinx2=25.
Finally, tanx2=2.
There is another way to do this problem as well. First we can find tanx2 by using the identity tanx=2tanx21tan2x2. And then we can find sinx2 and cosx2 from tanx2.

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