# trig identities topic assessment (integral maths) Q5 prove that 4[sin^4 (x) + cos^4 (x)] cos4x + 3

any ideas?? What have you tried?

If you can't prove from left to right, then try right to left (expanding is straightforward albeit messier). Solving RHS:

i got cos^4 (x) -6[sin^2 (x)cos^2 (x)] + sin^4 (x)

i do not know what to do next.. as i want to factorise a 4 out but i do not have any 4s.
also i need to add a 3 from the +3 part ^^ Original post by bobinabr23
Solving RHS:

i got cos^4 (x) -6[sin^2 (x)cos^2 (x)] + sin^4 (x)

i do not know what to do next.. as i want to factorise a 4 out but i do not have any 4s.
also i need to add a 3 from the +3 part ^^

3cos^4(x) + 3sin^4(x) - 3cos^4(x) - 3sin^4(x)
then pick out the LHS and argue then rest gives you 3?

Or you could split the 6cos^2sin^2 term up into two lots of 3sin^2cos^2 and use the usual pythagorean identity on each to get to the answer.
(edited 3 weeks ago) Original post by mqb2766
3cos^4(x) + 3sin^4(x) - 3cos^4(x) - 3sin^4(x)
then pick out the LHS and argue then rest gives you 3?

Or you could split the 6cos^2sin^2 term up into two lots of 3sin^2cos^2 and use the usual pythagorean identity on each to get to the answer.

ohh icic

for working LHS: our teacher told us to use the cos double angle formula which would give us in the form cos(2a).
do some exploiting till you get a 1/4 and then the 4 on the outer bracket would make it 1.

simplifying it further will give u +3 Original post by bobinabr23
ohh icic

for working LHS: our teacher told us to use the cos double angle formula which would give us in the form cos(2a).
do some exploiting till you get a 1/4 and then the 4 on the outer bracket would make it 1.

simplifying it further will give u +3

There are a few ways you could go and its sometimes worth doing the same question a few different ways. Here, rather than dropping the 3 from the working you could have guessed that it might drop out of the pythagorean identity and replaced it with
3 = 3*1 = 3(cos^2(x)+sin^2(x))
3 = 3*1^2 = 3*(cos^2(x)+sin^2(x))^2
The second form would be the reverse of "adding 0" and the first form would be be combined with the 3cos^2sin^2 + 3cos^2sin^2. Both give the extra two quartic terms as youd expect.