Trig question (1 Viewer)

[Estel]

It'll prolly be easy enough but I can't see it :S

In ABC acosA=bcosB
Prove using the cosine rule that the triangle is either isosceles or right angled.

Also if somebody has the proof for the expansion of tan3x using sinx and cosx it would be appreciated...

Thankyou

 

[Affinity]

me still trying to figure out where the cosine rule fits in.
mean while

acosA = bcosB
a/sinA = b/sinB
dividing first by second,

sinAcosA = sinBcosB
2sinAcosA = 2sinBcosB
sin2A = sin2B
sin2A - sin2B = 0
notice that 2A = (A+B) + (A-B) and 2B = (A+B) - (A-B)
so:
2sin(A-B)cos(A+B) = 0

either sin(A-B) = 0 or cos(A+B) = 0
A+B < 180 degrees, A,B < 180 degrees. &lt;--- Important
if sin(A-B) = 0 then A = B
if cos(A+B) = 0, A+B = 90 degrees and C = 90 degrees.


2.)
tan(3x) = {3sin(x) - 4[sin(x)]^3 } / {4[cos(x)]^3 - 3cos(x) }

 

[Estel]

Thanks a bunch

It's a question from Cambridge and they ask to use cosine rule, so well maybe there's a way that involves it. But meanwhile, I'll compute your method.

Part 2 the problem is changing it from the form you've got to the standard form.

 

[Estel]

I like your solution, I think I'm going to have to learn sums to products.

 

[Xayma]

acosA=a(b²+c²-a²)/2bc
bcosB=b(a²+c²-b²)/2ac

if acosA=bcosB
a(b²+c²-a²)/2bc=b(a²+c²-b²)/2ac

Cross multiplying

2a²c
(b²+c²-a²)=2b²c(a²+c²-b²)
a²b²+a²c²-a⁴=a²b²+b²c²-b⁴
a²c²-a⁴=b²c²-b⁴
c²(a²-b<sup>2</sup)=a⁴-b⁴
c²(a²-b²)=(a²+b²)(a²-b²)

Now if a=b (ie an isosceles triangle) both sides =0 therefore true for a=b

If not
c²=a²+b²
therefore ABC is a right angled triangle

 

[Affinity]

oh

anyway.
standard form.. you mean..
tan(3x)
= {3sin(x) - 4[sin(x)]^3 } / {4[cos(x)]^3 - 3cos(x) }
divide numerator and denominator by cos(x)^3..
to save writing, C= cos(x), S = sin(x) T = tan(x)

= [ 3T/C^2 - 4T^3 ] / [4 - 3/C^2]

now remember 1/C^2 = sec(x) ^ 2 = tan(x)^2 + 1

= [3T(T^2 + 1) - 4T^3] / [4 - 3(T^2 + 1)]

= [3T - T^3] / [ 1 - 3T^2 ]

 

[DcM]

Originally posted by Affinity
notice that 2A = (A+B) + (A-B) and 2B = (A+B) - (A-B)
so:
2sin(A-B)cos(A+B) = 0

thats no longer in Ext I....
but still good for Ext II

 

[CM_Tutor]

Originally posted by DcM
thats no longer in Ext I....
but still good for Ext II

You don't need it anyway. Affinity proved that sin 2A = sin 2B. From there, we proceed

Since A and B are both angles in a triangle (and, in fact, both must be acute), it follows that:

either 2A = 2B, in which case A = B, and the triangle is isosceles

or 2A = pi - 2B, in which case C = A + B = pi / 2, and the triangle is right-angled.

Converting sin 2A = sin 2B to 2sin(A - B)cos(A + B) = 0, as Affinity did, makes it easier to see these options, but it certainly isn't necessary