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###### trig identities topic assessment (integral maths) Q5

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3 weeks ago

prove that 4[sin^4 (x) + cos^4 (x)] ≡ cos4x + 3

any ideas??

any ideas??

Reply 1

3 weeks ago

What have you tried?

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

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

Reply 2

3 weeks ago

bobinabr23 OP

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 ^^

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 ^^

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 ^^

Looking towards the answer, can you "add 0" to this so

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)

Reply 4

2 weeks ago

bobinabr23 OP

Original post by mqb2766

Looking towards the answer, can you "add 0" to this so

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.

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

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.

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