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# C3 Question watch

1. Can anyone teach me how to do this question fully?
The mark scheme is missing so many steps
I cannot find R and alpha

2. (Original post by mystreet091234)

Can anyone teach me how to do this question fully?
The mark scheme is missing so many steps
I cannot find R and alpha

Ok, so I'm assuming you know how to use the product rule to get to e^-x(2cos2x-sin2x)?

Ok from there normally to find R, you square root a^2 and b^2so cosx=a and sinx=b

we have the form RCOS(X+alpha) this can also be written as acosx-bsinx

Using the equation we found before if we compare whats in the bracket to the version that we are given we can see that:

2cos2x-sinx2x
acos2x-bsin2x

so a=2 and b=1 (you don't need to worry about the signs here)

so to find R= square root 2^2 + 1^2
so we get square root of 5

Using what we know before we can see that cos2x=a which is 2
and sin2x=b which is 1

normally tanalpha= sin2x/cos2x (as tanx is sinx/cosx)

so we can see that tanalpha= 1/2

so inverse tan (1/2) we get 0.46

Does that help? It was kinda difficult to explain it on here,
Ok, so I'm assuming you know how to use the product rule to get to e^-x(2cos2x-sin2x)?

Ok from there normally to find R, you square root a^2 and b^2so cosx=a and sinx=b

we have the form RCOS(X+alpha) this can also be written as acosx-bsinx

Using the equation we found before if we compare whats in the bracket to the version that we are given we can see that:

2cos2x-sinx2x
acos2x-bsin2x

so a=2 and b=1 (you don't need to worry about the signs here)

so to find R= square root 2^2 + 1^2
so we get square root of 5

Using what we know before we can see that cos2x=a which is 2
and sin2x=b which is 1

normally tanalpha= sin2x/cos2x (as tanx is sinx/cosx)

so we can see that tanalpha= 1/2

so inverse tan (1/2) we get 0.46

Does that help? It was kinda difficult to explain it on here,
Yes!! it helps a lot thank you so much!

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