Question

Express :: 4 cos⁡θ+ 6 sin⁡θ in the form of
R sin⁡(θ+α),where R>0 and 0<α<1/2 π

please help :)

Answer 1

use the relation

R sin⁡(θ+α) = Rsinθcosα + Rcosθsinα

Answer 2

compare coefficients of cosθ and sinθ with the RHS of the above relation
for sin θ you will get
Rcosα = 6

do you see how i got that?

Answer 3

and what would Rsinα equal to?

Answer 4

what's an RHS ? unfortunately no :(

Answer 5

RHS = right hand side
Rsinθcosα + Rcosθsinα
Rcosα sinθ + Rsinα cosθ
compare: 6 sinθ + 4 cosθ

Answer 6

so

Rcosα = 6
and
Rsinα = 4

Answer 7

ok?

Answer 8

yeah , understood :D

Answer 9

ok now from the above 2 equations we can find the value of R and α as follows

Rsinα
------ = tanα = 2/3 - solve this to get α
Rcosα

find R

R^2(sin^2α + cos^2α) = 6^2 + 4^2
R^2 = 100

Answer 10

Why did you combine them by division ?

The mark scheme values are different ..

Answer 11

for α it's the same . but the R^2 it's = 4^2 + 6^2 which is weird

Answer 12

silly me 6^2 + 4^2 = 52 not 100!!!

Answer 13

i skipped a step
R^2cos^2α + R^2sin^2α = 4^2 + 6^2
factoring
R^2(cos^2α + sin^2α) = 4^2 + 6^2

Answer 14

combining by division eliminates R

Answer 15

yeah noticed it now :D
The second step , powering everything ..?

Answer 16

yes squaring each term
and sin^2α + cos^2α = 1

Answer 17

That's logically weird , Thanks so much and sorry such questions :D
Appreciate it :)

Answer 18

yw