How do I simplify this trigonometry expression?

[Miike012]

Homework Statement

Tan(15deg)/(1-Tan^2(15deg)


In my solution manual this is what they did...

2/2 * Tan(15deg)/(1-Tan^2(15deg)

= 1/2Tan(30deg)


Ok first off... I noticed they multiplyed by a fraction of one (2/2) which resulted in
2*tan(15) in the numerator... ( I did not know that you could multiply the coefficient infront by the deg in parenthesis to get tan(30)??)
Second: Where did the denominator go?

 

[Mark44]

Homework Statement

Tan(15deg)/(1-Tan^2(15deg)


In my solution manual this is what they did...

2/2 * Tan(15deg)/(1-Tan^2(15deg)

= 1/2Tan(30deg)


Ok first off... I noticed they multiplyed by a fraction of one (2/2) which resulted in
2*tan(15) in the numerator... ( I did not know that you could multiply the coefficient infront by the deg in parenthesis to get tan(30)??)
Second: Where did the denominator go?

Your title is misleading. From what I can see you are not simplifying a trig identity - you are simplifying a trig expression. When you prove an identity, you show that the two sides of a given equation are equal for all values of the variable. When you simplify an expression, you write it in simpler terms, possibly using identities.

And that's right, you can't just take a coefficient from outside a trig expression and move it into the angle. There was something else they did to get to that expression.

I would start with your original expression -- Tan(15deg)/(1-Tan^2(15deg) -- and rewrite it in terms of sin and cos.

 

[Miike012]

sin(15)/cos(15)/(1-sin^2(15)/cos^2(15))
= cos(15)sin(15)/( Cos^2(15) - Sin^2(15) )

 

[Miike012]

Still don't know why they decided to multiply the expression by 2/2...?

 

[Mark44]

Do you know an identity for cos²(x) - sin²(x)?

 

[Mark44]

... and another one for sin(x)cos(x)?

 

[Miike012]

I know the one for cos2(x) - sin2(x)?
But I am unsure of the one for sin(x)cos(x) is it similar to 2sin(x)cos(x)?

 

[jhae2.718]

What can you multiply sin(x)cos(x) by to write it with a 2sin(x)cos(x)?

 

[Miike012]

2...

 

[jhae2.718]

2/2, so you don't modify the value. Then you can use the 2sin(x)cos(x) identity on (2sin(x)cos(x))/2.

 

[Miike012]

Ok.. So I have 2/2* cos(15)*sin(15)/cos(30)
( cos^2(15) - Sin^2(15) = cos(30) I think? )

= sin30/2*cos30...
O still don't see how the denominator disapeared?

 

[eumyang]

Why can't you just use the tangent of a double angle formula?
$$\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$$

Or has the OP not learned it yet?

 

[jhae2.718]

What is sin(30)/cos(30) ?

 

[jhae2.718]

Why can't you just use the tangent of a double angle formula?
$$\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$$

Or has the OP not learned it yet?

Can't speak for anyone but myself, but maybe it's because I don't remember that one?

 

[Miike012]

Why can't you just use the tangent of a double angle formula?
$$\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$$

Or has the OP not learned it yet?

Would a 2 Have to be in the numerator for me to be able to use that identity though?

 

[Miike012]

Nevermind... I would just mult by 2/2 again then I would have Tan2Theta/2... I think?

 

[eumyang]

Would a 2 Have to be in the numerator for me to be able to use that identity though?

Um, yes, the 2 in the numerator is part of the formula.