Please solve q 3?

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2 Answers
May 4, 2018

The answer is 10, or D.

Explanation:

You can see that

#sin^2(5˚) + sin^2(85˚)# can be written as

#sin^2alpha + sin^2(90 - alpha)#. This can be simplified as follows:

#sin^2alpha + (sin90cosalpha - cos90sinalpha)^2 = sin^2alpha + cos^2alpha = 1#

Now we examine our series:

#sin^2(5˚) + sin^2(10˚) + sin^2(15˚) + ... + sin^2(75˚) + sin^2(80˚) + sin(85˚) + sin^2(90˚)#

Each term cancels out (#sin^2(5˚) + sin^2(85˚) = 1#, #sin^2(75˚) + sin^2(15˚) = 1# etc).

So we now make a table:

#0, 90#
#5, 85#
#10, 80#
#15, 75#
#20, 70#
#25, 65#
#30, 60#
#35, 55#
#40, 50#
#45, 45#

Each of these pairs have a value of #1# when the square of their sines is added.

Thus the sum is #1 * 10 = 10#.

Hopefully this helps!

May 4, 2018

#10#

Explanation:

#sin^2 5+sin^2 10+sin^2 15+....sin^2 85+sin^2 90--(*)#

we need to remember that

#(1) " "sintheta=cos(90-theta)#

#(2)" " sin^2theta+cos^2theta=1#

so

#sin85=cos5#
#sin10=cos80#
etc.

#(8)rarr sin^2 5+sin^2 10+sin^2 15+...cos^2 15+cos^2 10 +cos^2 5 +sin^2 90#

#(sin^2 5+cos^2 5)+(sin^2 10 +cos^2 10)+ ...+(sin^2 45+ cos^2 45)+sin^2 90#

#=1+1+....1+1#

#=9+1=10#
.