It wasted lot of time...

[lihinii]

I wasted too much time on the second part of this question

cos x + cos y = 1/3
sin x + sin y = 1/4 . Show that tan1/2( x + y) = 3/4. (Well this is ok but the next part is the problem)

Find the values of x and y.

2 cos( x+y )/2. cos( x-y )/2 =1/3
2 sin (x+ y)/2. cos(x-y )/2 = 1/4 gets you the first part.

can someone help me with the second part

thanx i.e.

 

[mmm4444bot]

tan1/2( x + y)

I'm not familiar with this notation. Tangent is a function; can you use function notation, instead ?



2 cos( x+y )/2. cos( x-y )/2 =1/3

2 sin (x+ y)/2. cos(x-y )/2 = 1/4

Please tell me what the periods are supposed to mean within each of these two equations.

 

[Deleted member 4993]

I wasted too much time on the second part of this question

cos x + cos y = 1/3
sin x + sin y = 1/4 . Show that tan1/2( x + y) = 3/4. (Well this is ok but the next part is the problem)

Find the values of x and y.

2 cos( x+y )/2. cos( x-y )/2 =1/3 .........................................(1)
2 sin (x+ y)/2. cos(x-y )/2 = 1/4 .......................................(2)

Divide (2) by (1)


gets you the first part.

can someone help me with the second part

thanx i.e.

 

[lihinii]

Sorry for the vagueness of the question

If cos x + cos y = 1/3 , and
sin x + sin y = 1/4

Show that
tan [1/2( x + y )] = 3/4.

Find all x and y. 0 < x,y < 360

Thanks.

 

[Deleted member 4993]

Sorry for the vagueness of the question

If cos x + cos y = 1/3 , and
sin x + sin y = 1/4

Show that
tan [1/2( x + y )] = 3/4.

Find all x and y. 0 < x,y < 360

Thanks.

Did you follow my instruction above?

 

[BigGlenntheHeavy]

\(\displaystyle One \ way, \ to \ wit:\)

\(\displaystyle cos(x)+cos(y) \ = \ \frac{1}{3} \ = \ 2cos\bigg(\frac{x+y}{2}\bigg)cos\bigg(\frac{x-y}{2}\bigg), \ (1)\)

\(\displaystyle sin(x)+sin(y) \ = \ \frac{1}{4} \ = \ 2sin\bigg(\frac{x+y}{2}\bigg)cos\bigg(\frac{x-y}{2}\bigg), \ (2)\)

\(\displaystyle Now, \ from \ (1), \ we \ have \ cos\bigg(\frac{x-y}{2}\bigg) \ = \ \frac{1}{6cos\bigg(\frac{x+y}{2}\bigg)}\)

\(\displaystyle Hence, \ subbing \ into \ (2), \ \frac{2sin\bigg(\frac{x+y}{2}\bigg)}{6cos\bigg(\frac{x+y}{2}\bigg)} \ = \ \frac{1}{4}, \ \implies \ \frac{tan\bigg(\frac{x+y}{2}\bigg)}{3} \ = \ \frac{1}{4}\)

\(\displaystyle Ergo, \ tan\bigg(\frac{x+y}{2}\bigg) \ = \ \frac{3}{4}, \ QED.\)