Oxford University MAT 2021 Solved Papers

© examsiri.com

Question : 7 of 10

Marks: +1, -0

The sum

\sin ^2(1^{∘})+\sin ^2(2^{∘})+\sin ^2(3^{∘})+⋯+\sin ^2(89^{∘})+\sin ^2(90^{∘})

is equal to

44 ,
$44 \;{1}/{2}$ ,
45 ,
$45 \;{1}/{2}$ ,
46 .

Solution:

We can use the fact that

\sin ^2(90^{∘}-n)=\cos ^2(n)

for any

n

, and

\sin ^2 x+\cos ^2 x=1

. So we have

\sin ^2 1^{∘}+\sin ^2 89^{∘}=1

and

\sin ^2 2^{∘}+\sin ^2 88^{∘}=1

and so on up to

\sin ^2 44^{∘}+\sin ^2 46^{∘}=1

. We also have

\sin ^2 45^{∘}=\;{1}/{2}

and

\sin ^2 90^{∘}=1

for a total of

45 \;{1}/{2}

© examsiri.com

Go to Question:

Prev Question

Next Question