Inverse Trigonometry question

[Saitama]

Homework Statement

cos(2tan^-1(1/7)) equals-

a)sin(4cot^-13)
b)sin(3cot^-14)
c)cos(3cot^-14)
d)sin(4cot^-14)

Homework Equations

The Attempt at a Solution

I am completely stuck in this one. I tried using the formula $2tan^{-1}x=tan^{-1}(\frac{2x}{1-x^2})$. Using this formula i got $cos(tan^{-1}\frac{7}{24})$.
Now i am stuck, i can't think of anything else on how to proceed.
Any help is appreciated. :)

 

[Curious3141]

Homework Statement

cos(2tan^-1(1/7)) equals-

a)sin(4cot^-13)
b)sin(3cot^-14)
c)cos(3cot^-14)
d)sin(4cot^-14)

Homework Equations

The Attempt at a Solution

I am completely stuck in this one. I tried using the formula $2tan^{-1}x=tan^{-1}(\frac{2x}{1-x^2})$. Using this formula i got $cos(tan^{-1}\frac{7}{24})$.
Now i am stuck, i can't think of anything else on how to proceed.
Any help is appreciated. :)

I worked out that it's equal to choice a), but it wasn't really quick. I basically had to calculate the exact value ($\frac{24}{25}$), then see which of the choices were likely candidates. It's unlikely to be the ones with the form cos or sine of a triple angle, as they involve cubes of square root surds. So I worked out the first one of the form sin4θ, and got lucky, because it was also equal to $\frac{24}{25}$.

To work out the exact value in each case, I used the double-angle formulae directly (twice for the sin4θ form). Then I drew the right triangle to work out the individual sines and cosines.

I'll get back to you if I find a quicker way.

 

[Saitama]

I worked out that it's equal to choice a), but it wasn't really quick. I basically had to calculate the exact value ($\frac{24}{25}$), then see which of the choices were likely candidates. It's unlikely to be the ones with the form cos or sine of a triple angle, as they involve cubes of square root surds. So I worked out the first one of the form sin4θ, and got lucky, because it was also equal to $\frac{24}{25}$.

To work out the exact value in each case, I used the double-angle formulae directly (twice for the sin4θ form). Then I drew the right triangle to work out the individual sines and cosines.

I'll get back to you if I find a quicker way.

Thanks for the reply Curious! :)

I too thought of evaluating all the options and see if they matches my answer. But then it would take a lot of time. The answer in the key is (a).

 

[I like Serena]

My first approach was the same.
It works, but it is quite a bit of work.

Alternatively, you can also draw the triangles on scale, and measure what you need.
This option tends to be forgotten by mathematicians, but it would typically be a first choice of an engineer.

As it is, a rough sketch is enough to see (reasonably) quickly that the other 3 answers don't fit.

 

[ehild]

Can not you use a calculator?
As Curious said, it is easy to get that cos(2tan^-1(1/7))=24/25.
(With the notation x=tan^-1(1/7)

$$cos(2x)=\frac{1-\tan^2(x)}{1+\tan^2(x)}=\frac{1-1/49}{1+1/49}$$
)
To calculate a,b,c,d, replace
cot^-1(x) by tan^-1(1/x). Use the addition low of sine and cosine, express sin(x) and cos(x) and sin(2x) and cos(2x) by tan(x). The triple-angle expression contain square roots, they are irrational so they can not be equal to 24/25.

a.) and d.) can be written in the form

$$\sin(4x)=2 \sin(2x)\cos(2x)=2\frac{2 \tan(x)}{1+\tan^2(x)} \frac{1-\tan^2(x)}{1+\tan^2(x)}$$

where x=tan^-1(1/3) for a.) and x=tan^-1(1/4) for d.)

Use that tan(tan^-1(x))=x.

ehild

 

[Saitama]

Thank you both ILS and ehild for the help!

I am currently studying Inverse trigonometry and my teacher gave a booklet of many questions. All the questions were getting solved by the manipulation of identities and formulas so i thought this question should also get solved without using a calculator. Well, if its not possible to solve it without calculator, i will leave it. I already got my answer by using the calculator but i thought if i could do it without calculator, i posted the question here to get some help.

Thanks for the help! :)

 

[ehild]

Post #5 shows how to do it without calculator.

ehild

 

[Whovian]

Here's a nice solution.

Draw a right triangle. Use the right-triangle definitions of sin, cos, tan, cot, sec, an csc. Make the right triangle so you can, um, find what you're looking for. I don't really know how to explain this properly.

 

[piknless]

here is a solution I thought of on the whim. Using the double angle property, cos(2tan-1(1/7)) turns into cos(2x) where x equals tans-1 (1/7), which is ((cos^2)tan-1(1/7))-((sin^2)tan-1(1/7)). now you know that the tangent of the inverse tangent of 1/7 equals 1/7. so draw a triangle with angle inverse tan 1/7. you get cos of inverse tan equals 7/root 50. do the same for sin. Then plug it back in the original formula, and you get (49/50)-(1/50)=(48/50)=(24/25)

 

[piknless]

starting from where you left off, cos(tan-1 (7/24)) draw a triangle with angle tan inverse 7/24. The tangent of that angle would be 7/24. by using pythagorean theorem we can deduce the last side is 25. So the cosine of the inverse tangent of (7/24) is equivalent to 24/25.

 

[HallsofIvy]

here is a solution I thought of on the whim. Using the double angle property, cos(2tan-1(1/7)) turns into cos(2x) where x equals tans-1 (1/7), which is ((cos^2)tan-1(1/7))-((sin^2)tan-1(1/7)). now you know that the tangent of the inverse tangent of 1/7 equals 1/7. so draw a triangle with angle inverse tan 1/7. you get cos of inverse tan equals 7/root 50. do the same for sin. Then plug it back in the original formula, and you get (49/50)-(1/50)=(48/50)=(24/25)

Hardly a "whim"! That's a very good solution.

 

[Bohrok]

That approach you used piknless was the first one that came to my mind I started typing it out until I realized other posters showed how they got 24/25, and that still doesn't tell you which of the four choices is the same as 24/25...

I'm still trying to see if there's a way to derive the answer from the given expression without finding the exact answer and seeing which one is equal to it.

 

[Curious3141]

Thank you both ILS and ehild for the help!

I am currently studying Inverse trigonometry and my teacher gave a booklet of many questions. All the questions were getting solved by the manipulation of identities and formulas so i thought this question should also get solved without using a calculator. Well, if its not possible to solve it without calculator, i will leave it. I already got my answer by using the calculator but i thought if i could do it without calculator, i posted the question here to get some help.

Thanks for the help! :)

I didn't use a calculator at all. Sorry if that wasn't clear. I used the double-angle identities and a sketch of the right triangle (and Pythagoras' Theorem) for the calculation in each case. The problem is that we need to calculate the choices this way too. As I said, we can quickly eliminate the triple-angle forms, which leaves only a) and d) to test, and the first choice gives the answer immediately.

The jury's still out on whether there's an ingenious way to just "see" it without calculation of each choice, but so far, it seems negative from all the posts here.