A maths questions

anpanman SG

1. Prove that tan(pi/4 +x) = (1 + sin2x) / cos 2x

2. When the function f(x) = 2x^3 + ax^2 + bx + 6 is divided by x^2 + x - 2, the remainder is -8x +4. Find the value of a and b.

a is 1 and b is -13

By using a suitable substitution, or otherwwise, solve the equation

6y^3 - 13y^2 +y +2 = 0

Well, if I want to use a suitable sub, then how can i link this question to the above equation of 2x^3 + x^2 - 13x + 6?

The powers are all in random order. Thanks!

TrueHeart SG

1. Prove that tan(pi/4 +x) = (1 + sin2x) / cos 2x

2. When the function f(x) = 2x^3 + ax^2 + bx + 6 is divided by x^2 + x - 2, the remainder is -8x +4. Find the value of a and b.

a is 1 and b is -13

By using a suitable substitution, or otherwwise, solve the equation

6y^3 - 13y^2 +y +2 = 0

Well, if I want to use a suitable sub, then how can i link this question to the above equation of 2x^3 + x^2 - 13x + 6?

The powers are all in random order. Thanks!

For the first question,

R.H.S.

= (1+sin2x)/cos2x

= sin^2x + cos^2x + 2sinxcosx / cos^2x - sin^2x

= (cosx+sinx)^2 / (cosx+sinx)(cosx-sinx)

= (cosx+sinx) / (cosx-sinx)

= [(sqrt 2)sin(x+pi/4)] / [(sqrt2)sin(pi/4-x)]

= sin(x+pi/4) / cos(x+pi/4)

= tan(pi/4+x)

= L.H.S.

For the second question, let x =1/y

Solve for 2x^3+x^2-13x+6 = 0.

Once you get the three values of x, sub x = 1/y in to find the three roots of 6y^3-13y^2+y+2 = 0

309795628 SG

！！！看不懂哦为什么都在说英文额。。。

anpanman SG

For the first question,

R.H.S.

= (1+sin2x)/cos2x

= sin^2x + cos^2x + 2sinxcosx / cos^2x - sin^2x

= (cosx+sinx)^2 / (cosx+sinx)(cosx-sinx)

= (cosx+sinx) / (cosx-sinx)

= [(sqrt 2)sin(x+pi/4)] / [(sqrt2)sin(pi/4-x)]

= sin(x+pi/4) / cos(x+pi/4)

= tan(pi/4+x)

= L.H.S.

For the second question, let x =1/y

Solve for 2x^3+x^2-13x+6 = 0.

Once you get the three values of x, sub x = 1/y in to find the three roots of 6y^3-13y^2+y+2 = 0

= (cosx+sinx) / (cosx-sinx)

= [(sqrt 2)sin(x+pi/4)] / [(sqrt2)sin(pi/4-x)]

Don't understand how you can convert the above part...i suppose the highlighte dpart should be cos anyway, even if that's the case, how did you go on to get sin(x+pi/4) / cos(x+pi/4)?

TrueHeart SG

= (cosx+sinx) / (cosx-sinx)

= [(sqrt 2)sin(x+pi/4)] / [(sqrt2)sin(pi/4-x)]

Don't understand how you can convert the above part...i suppose the highlighte dpart should be cos anyway, even if that's the case, how did you go on to get sin(x+pi/4) / cos(x+pi/4)?

Sorry, I skipped some steps. Let me explain them part by part.

Since you will want to introduce pi/4 in,

cosx+sinx will have to become (sqrt2)sin(pi/4)cosx + (sqrt2)cos(pi/4)sinx. Note that sin(pi/4)=cos(pi/4)=1/(sqrt2)

Taking out (sqrt2) and you will have (sqrt2)[sin(pi/4)cosx + cos(pi/4)sinx],

which can be simplified into (sqrt2)[sin(x+pi/4)].

The same goes for cosx - sinx, and the end result is [(sqrt2)sin(pi/4-x)] alright.

The two (sqrt2) will cancel each other out, whereas sin(pi/4 - x) can be converted to cos(x + pi/4).

wee_ws SG

Hi,

cos x + sin x

= sin x + cos x

= sqrt(2) sin (x + pi/4), due to the application of the R-formula.

Similarly, cos x - sin x = sqrt(2) cos (x + pi/4).

The quotient of the two expressions gives tan (x + pi/4).

Remember these results:

a cos x + b sin x = sqrt(a^2 + b^2) cos (x - alpha), alpha = arctan(b/a),

a sin x + b cos x = sqrt(a^2 + b^2) sin (x + alpha), alpha = arctan(b/a).

Thanks!

Cheers,
Wen Shih

Lee012lee SG

Another Method

Question

1 + sin 2x

Prove that tan (π/4 +x) = -------------

cos 2x

Solution

1 + sin 2x

RHS = ------------

cos 2x

sin² x + cos² x + 2 sin x cos x

= ----------------------------------

cos² x – sin² x

(cos x + sin x)²

= ------------------------------------

(cos x + sin x) (cos x – sin x )

cos x + sin x

= ----------------

cos x – sin x

Divide the numerator and denominator by cos x

cos x + sin x

---------------

cos x

= -----------------

cos x – sinx

--------------

cos x

1 + tan x

= -----------

1 - tan x

1 + tan x

= -----------

1 – (1)tan x

Since tan (π/4 ) = 1

tan (π/4) + tan x

= ---------------------

1 - tan (π/4) tan x

= tan (π/4 +x)

Hence, LHS = RHS (Proved)

wee_ws SG

Hi Lee012lee,

Cool! That's the nice thing about maths in its flexibility of reaching the same outcome. Thanks!

Cheers,
Wen Shih

eagle SG

I believe it would be easier for most students from LHS to RHS, also using Lee's method (in reverse)

Reason being, it's always easier to work from tan to sin and cos because tan is a combination of both ;)

anpanman SG

With regards to question 2,I still don't get what TrueHeart is trying to say. (sorry, forgot about this question i posted at the later part of the post until today so didnt clarify my doubts about this question)

Thanks everyone for helping.

TrueHeart SG

Thanks everyone for helping.

Hi,

which part do you have difficulties in understanding? Sorry, but could you please kindly point them out to me?

anpanman SG

Hi,

which part do you have difficulties in understanding? Sorry, but could you please kindly point them out to me?

why is it that we must equate x = 1/y?

Thanks!

TrueHeart SG

why is it that we must equate x = 1/y?

Thanks!

First, let us make 2x^3+x^2-13x+6 be equal to zero. By subbing in x=(1/y), and after some simplifying, we will have 6y^3-13y^2+y+2=0 Thus, the link between x and y is that x = 1/y.

Hence, after finding the roots of x, we can simply sub x = 1/y in to find the roots of the second equation. Feel free to post anymore queries which you may have, I will try and see if I can help out.

anpanman SG

First, let us make 2x^3+x^2-13x+6 be equal to zero. By subbing in x=(1/y), and after some simplifying, we will have 6y^3-13y^2+y+2=0 Thus, the link between x and y is that x = 1/y.

Hence, after finding the roots of x, we can simply sub x = 1/y in to find the roots of the second equation. Feel free to post anymore queries which you may have, I will try and see if I can help out.

Ok, fully understood. Thank you!