Solving Trig Problems: Finding x for (sinx)^4 & (cosx)^4

[armolinasf]

Homework Statement

I have to trig problems which I'm not entirely sure how to approach:


1) For which angles x is (sinx)^4-(cosx)^4 > (sinx)^2-(cosx)^2?

2) For what value x is the expression (sinx-cosx)^2+(sinx+cosx)^2 as large as possible?

The Attempt at a Solution

For the first one I was able to add and subtract 1 to the left side which gave me a difference of squares that let me use the pythagorean identity leaving me with sinx^2-cosx^2>sinx^2_cosx^2 But this doesn't really make sense, so I'm not sure where to go.

For the second, am I supposed to do this numerically? or how would I approach it analytically?

Thanks for the help.

 

[stevenb]

Homework Statement

I have to trig problems which I'm not entirely sure how to approach:


1) For which angles x is (sinx)^4-(cosx)^4 > (sinx)^2-(cosx)^2?

2) For what value x is the expression (sinx-cosx)^2+(sinx+cosx)^2 as large as possible?

The Attempt at a Solution

For the first one I was able to add and subtract 1 to the left side which gave me a difference of squares that let me use the pythagorean identity leaving me with sinx^2-cosx^2>sinx^2_cosx^2 But this doesn't really make sense, so I'm not sure where to go.

For the second, am I supposed to do this numerically? or how would I approach it analytically?

Thanks for the help.

On the first one, you are missing a very simple approach. What does (sinx)^2-(cosx)^2 times (sinx)^2+(cosx)^2 equal? Since (sinx)^2+(cosx)^2 is equal to one, you can multiply the right hand side by it without changing anything.

For the second one, take the derivative and set it equal to zero. The roots are potentially points of maximum value. You can solve the equation numerically using Newton's method or another approach. Also, plot out the function to get a visual feel for what is happening.

 

[armolinasf]

Great. That definitely points me in the right direction, Thanks.

 

[stevenb]

Great. That definitely points me in the right direction, Thanks.

Also, take a good look at the second problem. It looks to me there is an easier way to solve this one without numerical solution.

 

[armolinasf]

for the first one you get sin^2x-cos^2x>1 I'm tempted to use a pythagorean identity by subtracting sin^2x and then dividing both sides by cos^2x but that would give me something which doesn't make sense: -1>1.

Do I have to get x isolated on one side to figure out which value makes the statement true?

For b) I expanded the binomials and it simplified to 2(sin^2x+cos^2x)=2*1 so is 2 the greatest possible value? and why does this work?

 

[SammyS]

for the first one you get sin^2x-cos^2x>1 I'm tempted to use a pythagorean identity by subtracting sin^2x and then dividing both sides by cos^2x but that would give me something which doesn't make sense: -1>1.

Do I have to get x isolated on one side to figure out which value makes the statement true?

For b) I expanded the binomials and it simplified to 2(sin^2x+cos^2x)=2*1 so is 2 the greatest possible value? and why does this work?

That's right for part (b).

For part (a): Use the fact that A⁴ ‒ B⁴ = (A² ‒ B²)(A² + B²).

 

[armolinasf]

using difference of squares I get (sin^2x-cos^2x)(sin^2x+cos^2x)>(sin^2x-cos^2x) divide both sides by sin^2x-cos^2 and using a pythagorean identity you get 1>1 correct?

 

[stevenb]

using difference of squares I get (sin^2x-cos^2x)(sin^2x+cos^2x)>(sin^2x-cos^2x) divide both sides by sin^2x-cos^2 and using a pythagorean identity you get 1>1 correct?

Yes. They gave you a trick question. The relation is an identity, so one side can not the greater that the other. Hence, there is no angle that meets the criterion.

The second problem is also a trick question.