...with the cube root in this equation tan just remains positive??

I know that square roots can are +/- but does the same not apply to cube roots? :frown:

Sorry for the ridiculously silly questions

Reply 42

Zacken

22

Original post by Philip-flop

...

If the original question was

\cos^2 \theta = \frac{3}{4}

then you get

\cos \theta = \pm \sqrt{\frac{3}{4}}

so you get two different trig equations to solve.

But with

]tan^3 \theta = \frac{1}{8}

, cube rooting gives only

\tan \theta = + \sqrt[3]{\frac{1}{8}}

which is only one trig equation to solve. I'm not really sure what you're asking tbh, could you post the original question and your working out?

Original post by Philip-flop

I know that square roots can are +/- but does the same not apply to cube roots? :frown:

Sorry for the ridiculously silly questions

Nope.

Take -1 for instance. Squaring it, or raising it to any even power, will give the result of +1. This is also the case for the number +1 itself, if you raise it to any even power.

If you were to cube -1, then you would be multiplying it by itself 3 times and you come back to -1 again. Same with any odd power. Now if you apply odd powers to the number +1, then you would just come back to +1 again.

So; for even powers, all numbers go positive; hence +/- solutions for any equations with square roots. For odd powers, negatives stay negative, and positives stay positive. Therefore odd roots of a negative will be negative, and odd roots of a positive are positive. Hence no +/- in the answer.

Reply 44

Philip-flop

OP

18

Original post by Zacken

could you post the original question and your working out?

Taken from the Edexcel C3 Modular Maths Textbook - Exercise 6C Questions: 5(e) and 5(g)

Solve, for the values of

\theta

in the interval

0 \leq \theta \leq 360

...

Q5(e)

Q5(g)

Attachment not found

OP

Original post by RDKGames

Nope.

Take -1 for instance. Squaring it, or raising it to any even power, will give the result of +1. This is also the case for the number +1 itself, if you raise it to any even power.

If you were to cube -1, then you would be multiplying it by itself 3 times and you come back to -1 again. Same with any odd power. Now if you apply odd powers to the number +1, then you would just come back to +1 again.

So; for even powers, all numbers go positive; hence +/- solutions for any equations with square roots. For odd powers, negatives stay negative, and positives stay positive. Therefore odd roots of a negative will be negative, and odd roots of a positive are positive. Hence no +/- in the answer.

Oh yeah!! That's it! Thank you so much. I don't know why but I seem to know these things already but when it comes to applying it to questions I panic and forget simple things like that.

Great explanation. It makes perfect sense now.

Original post by Philip-flop

Taken from the Edexcel C3 Modular Maths Textbook - Exercise 6C Questions: 5(e) and 5(g)

Solve, for the values of

\theta

in the interval

0 \leq \theta \leq 360

A bit messy.

So you have

\sec^2\theta = \frac{4}{3}

so

\cos^2\theta = \frac{3}{4}

since you take reciprocals of both sides.

At which point

\cos \theta = \pm \frac{\sqrt3}{2}

Your tan looks good but it would just be easier to rewrite cube root of 1/8 as 1/2.

Also I think learning and remembering the general solutions of trigonometric functions in FP1 can make it so much neater for you to solve these types in C3. It's a really short topic, look at it if you can. I feel like the quadrant thing just leaves much more room for error.

(edited 7 years ago)

Reply 47

ValerieKR

10

Original post by RDKGames

Your tan looks good

Reply 48

Philip-flop

OP

18

Original post by RDKGames

A bit messy.

So you have

\sec^2\theta = \frac{4}{3}

so

\cos^2\theta = \frac{3}{4}

since you take reciprocals of both sides.

At which point

\cos \theta = \pm \frac{\sqrt3}{2}

Your tan looks good but it would just be easier to rewrite cube root of 1/8 as 1/2.

Also I think learning and remembering the general solutions of trigonometric functions in FP1 can make it so much neater for you to solve these types in C3. It's a really short topic, look at it if you can. I feel like the quadrant thing just leaves much more room for error.