Help Simplifying a Trig Expression

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  • #1
opus
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Homework Statement


Simplify: $$sin^2\left(x\right)+sin^2\left(x\right)tan^2\left(x\right)$$

Homework Equations


All trigonometric identities.

The Attempt at a Solution


(i) ##sin^2\left(x\right)+sin^2\left(x\right)tan^2\left(x\right)##

(ii) ##sin^2\left(x\right)+sin^2\left(x\right)⋅\left(\frac{sin^2\left(x\right)}{cos^2\left(x\right)}\right)##

(iii) Multiply both terms by ##\frac{1}{sin^2\left(x\right)}##

(iv) ##1+\frac{sin^2\left(x\right)}{cos^2\left(x\right)}##

(v) ##1+tan^2\left(x\right)##

(vi) ##sec^2\left(x\right)##

Apologies for it looking a little sloppy. Spend some time trying to get the parentheses inside parentheses but kept getting errors.
Would anyone mind pointing out a spot where I made an error?
 

Answers and Replies

  • #2
Charles Link
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You can factor out ## \sin^2(x) ##. You don't just mulriply by ## \frac{1}{\sin^2(x) } ##. ## \\ ## Otherwise you got it correct. You can then take ## \sin^2(x) ## times ## \sec^2(x) ## to finish it up, and get ## \frac{\sin^2(x)}{\cos^2(x)} ## that you should recognize as something else.
 
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  • #3
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Ah dang! Should've seen that one.
For my knowledge, why can't I multiply out like I did? I figured since I did it to both terms, it was an allowable thing to do.
 
  • #4
Charles Link
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Ah dang! Should've seen that one.
For my knowledge, why can't I multiply out like I did? I figured since I did it to both terms, it was an allowable thing to do.
Ignoring the problem of division by zero, you can multiply by ## 1=\frac{\sin^2(x)}{\sin^2(x)} ##, if you don't want to factor out ## \sin^2(x) ##. ## \\ ## Multiplying by ## 1 ## doesn't change the expression. Multiplying by anything else does.
 
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  • #5
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So you're saying, it is not acceptable to multiply both terms by anything that does not have a value of 1, otherwise it will change the value of the whole expression? Also didn't think about the possibility of ##sin^2\left(x\right)## being equal to zero so that would cause problems.
 
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  • #6
opus
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Let me ask one more thing in regards to keeping the values of an expression even after manipulation- is what you say true for expression only, or both expressions AND equations.
The reason I ask is because in some trigonometry derivations, such as how to get the sum of angles formula for cosine, we have at some point:
2-2cos(u+v) = 2-2cos(u)cos(v)+2sin(u)sin(v)
and the final step is to subtract two from both sides and divide both sides by 2 to get the sum of angles formula for cosine.

This is clearly allowed because if you do anything to one side of an equation you do it to the other and the equation remains true. Why is this not true for terms of an expression?

I guess this is a very elementary question, but I've never thought about it until now.
 
  • #7
Charles Link
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For an equation, it is like a balance for weighing things with equal weights on both sides. As long as you do the same thing to both sides of the balance, it stays balanced. ## \\ ## in this case you have ## Expression=\sin^2(x)+\sin^2(x) \tan^2(x) ##. You still basically have an equation, with ## Expression ## on one side (the left side), so that if you add 2 to the right side you must also subtract 2 to add a total of zero. Similarly, if you divide the right side by ## sin^2(x) ##, you must also multiply the right side by ## sin^2(x) ##, or the right side will change and will no longer equal ## Expression ##.
 
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  • #8
Ray Vickson
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Homework Statement


Simplify: $$sin^2\left(x\right)+sin^2\left(x\right)tan^2\left(x\right)$$

Homework Equations


All trigonometric identities.

The Attempt at a Solution


(i) ##sin^2\left(x\right)+sin^2\left(x\right)tan^2\left(x\right)##

(ii) ##sin^2\left(x\right)+sin^2\left(x\right)⋅\left(\frac{sin^2\left(x\right)}{cos^2\left(x\right)}\right)##

(iii) Multiply both terms by ##\frac{1}{sin^2\left(x\right)}##

(iv) ##1+\frac{sin^2\left(x\right)}{cos^2\left(x\right)}##

(v) ##1+tan^2\left(x\right)##

(vi) ##sec^2\left(x\right)##

Apologies for it looking a little sloppy. Spend some time trying to get the parentheses inside parentheses but kept getting errors.
Would anyone mind pointing out a spot where I made an error?
Factor out the ##\sin^2 x##:
$$\sin^2 x + \sin^2 x \:\tan^2 x = \sin^2 x \left( 1 + \frac{\sin^2 x}{\cos^2 x} \right) = \sin^2 x \frac{\cos^2 x + \sin^2 x}{\cos^2 x}.$$
Now simplify.
 
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  • #9
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So you're saying, it is not acceptable to multiply both terms by anything that does not have a value of 1, otherwise it will change the value of the whole expression?
Right.

There's a big difference between an equation (where you start off with two quantities being given as equal) and an expression such as what you have in this problem. With an expression, there are only a very few operations you can perform:
  • Add 0 to it.
  • Multiply it by 1 in some form.
  • Simplify the expression, such as by factoring it or expanding it by multiplying its factors.
The 0 that gets added might be in some more complex form, such as adding 2 and then subtracting 2, as described in post #7. Multiplying by 1 could also be a bit more complex, as you can multiply by <something> divided by itself, provided that <something> is not equal to zero.

If you start with an equation, there are many more things you can do, as long as you consistently apply the same operation to both sides.
 
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  • #10
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Homework Equations


All trigonometric identities.
Three important identities of those not listed are the so-called Pythagorean identities:

##\sin^2(x) + \cos^2(x) = 1##
##\tan^2(x) + 1 = \sec^2(x)##
##\csc^2(x) + 1 = \cot^2(x)##.

The one involving tan(x) and sec(x) can be used to good advantage in this problem, yielding a simplification about as quickly as Ray's suggestion.
 
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  • #11
opus
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Extremely helpful as always. Thanks for all the responses!
 
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