Question

if x=a cosθ - b sinθ and y=a sinθ + b cosθ, how would i show that x^2 +y^2=a^2+b^2

Answer 1

ugh trigonometry :/

Answer 2

@alexwee123 right? tell me about it. i was doing fine until i met this strand in my course.

Answer 3

get an expression for x^2
get an expression for y^2
add them up
collect common terms
use the fact that \(\cos^2(\theta)+\sin^2(\theta)=1\) and answer to pop out...

Answer 4

ya i always hated it but i still got an A on my trig test :D

Answer 5

@alexwee123 lucky.. ! i still have three more trig units after this -.-

Answer 6

for x^2 i got acos^2θ + bsin^2θ , but for y^2 i got asin^2θ + absinθcosθ + absinθcosθ + bcos^2θ ....

Answer 7

@asnaseer

Answer 8

your expressions are not correct, remember:\[(ab+cd)^2=a^2b^2+2abcd+c^2d^2\]and similarly:\[(ab-cd)^2=a^2b^2-2abcd+c^2d^2\]

Answer 9

so (asinθ)^2 would be a^2sin^2θ right?

Answer 10

yes

Answer 11

k i'll try it again..

Answer 12

ok - good luck :)

Answer 13

@asnaseer k so i did it and got \[x ^{2} = a^{2}\cos ^{2}(\theta)-2acos (\theta)bsin (\theta)+b ^{2}\sin ^{2}(\theta)\]\[y ^{2} = a^{2}\sin ^{2}(\theta)+2asin (\theta)bcos (\theta)+b ^{2}\cos ^{2}(\theta)\]

then i know that
\[[a^{2}\cos ^{2}(\theta)-2acos (\theta)bsin (\theta)+b ^{2}\sin ^{2}(\theta) ] + [a^{2}\cos ^{2}(\theta)-2acos (\theta)bsin (\theta)+b ^{2}\sin ^{2}(\theta)]\]\[= a^{2}\cos ^{2}(\theta)+b ^{2} \sin ^{2} (\theta)+a ^{2}\sin ^{2}(\theta) + b ^{2}\cos ^{2}\theta\]

... right?

Answer 14

perfect!

Answer 15

you just need to collect "like" terms now and you are done

Answer 16

try and collect terms such that you end up with this expression somewhere:\[(cos^2(\theta)+\sin^2(\theta))\]

Answer 17

because that expression always equals one

Answer 18

\[a^2\cos^2(θ)+b^2\cos^2(θ)+a^2\sin^2(θ)+b^2\sin^2(θ)\] ?

:S

Answer 19

there are two terms here that have \(a^2\) in them and two terms that have \(b^2\) in them.

Answer 20

try and make use of that fact

Answer 21

e.g.\[a^2p^2+a^2q^2=a^2(p^2+q^2)\]

Answer 22

\[a^2\cos^2(θ)+a^2\sin^2(θ)+b^2\sin^2(θ)+b^2\cos^2(θ)\]\[=a^2(\cos^2(θ)+\sin^2(θ))+b^2(\sin^2(θ)+\cos^2(θ))\]\[=a^2(1)+b^2(1)\]\[=a^2+b^2\]

AIOSHADS ! :D YEAH ?

Answer 23

you got it! well done! :D

Answer 24

yay ! thanks so much :)

Answer 25

yw - I'm glad I was able to help :)