Eliminate θ and ϕ between the following equations: {sin⁡θ+sin⁡ϕ=xcos⁡θ+cos⁡ϕ=ytan⁡θ2tan⁡ϕ2=z What Ive

By sum to product and product to sum formulas we have {sin⁡θ+sin⁡ϕ=2sin⁡(θ+ϕ2)cos⁡(θ−ϕ2)=xcos⁡θ+cos⁡ϕ=2cos⁡(θ+ϕ2)cos⁡(θ−ϕ2)=ytan⁡θ2tan⁡ϕ2=cos⁡(θ−ϕ2)−cos⁡(θ+ϕ2)cos⁡(θ−ϕ2)+cos⁡(θ+ϕ2)=z⇒{2ab=x2cb=yb−cb+c=z and since a2+c2=1 we obtain b=±12x2+y2 c=±yx2+y2 and then z=±12x2+y2∓yx2+y2±12x2+y2±yx2+y2=x2+y2−2yx2+y2+2y

Write u=tan⁡θ2,v=tan⁡ϕ2. Then we get {2(u+v)(1+uv)=x(1+u2)(1+v2)2−2u2v2=y(1+u2)(1+v2)uv=z Now setting s=u+v and using the third relation, we obtain {2s(1+z)=x(1−2z+z2+s2)2(1−z2)=y(1−2z+z2+s2) Dividing, we obtain the relation s=x(1−z)y, which youve

Hints Writing θ=2p,ϕ=2q We have sin⁡(p+q)x=cos⁡(p+q)y=±1x2+y2 Using this one can find cos⁡(p−q) Now use z−1z+1=…=cos⁡(p+q)cos⁡(p−q)

By sum to product and product to sum formulas we have {sin⁡θ+sin⁡ϕ=2sin⁡(θ+ϕ2)cos⁡(θ−ϕ2)=xcos⁡θ+cos⁡ϕ=2cos⁡(θ+ϕ2)cos⁡(θ−ϕ2)=ytan⁡θ2tan⁡ϕ2=cos⁡(θ−ϕ2)−cos⁡(θ+ϕ2)cos⁡(θ−ϕ2)+cos⁡(θ+ϕ2)=z⇒{2ab=x2cb=yb−cb+c=z and since a2+c2=1 we obtain b=±12x2+y2 c=±yx2+y2 and then z=±12x2+y2∓yx2+y2±12x2+y2±yx2+y2=x2+y2−2yx2+y2+2y

Write u=tan⁡θ2,v=tan⁡ϕ2. Then we get {2(u+v)(1+uv)=x(1+u2)(1+v2)2−2u2v2=y(1+u2)(1+v2)uv=z Now setting s=u+v and using the third relation, we obtain {2s(1+z)=x(1−2z+z2+s2)2(1−z2)=y(1−2z+z2+s2) Dividing, we obtain the relation s=x(1−z)y, which youve

Hints Writing θ=2p,ϕ=2q We have sin⁡(p+q)x=cos⁡(p+q)y=±1x2+y2 Using this one can find cos⁡(p−q) Now use z−1z+1=…=cos⁡(p+q)cos⁡(p−q)

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High School
Geometry
Trigonometry
Trigonometric equations and identities

Marla Payton

Answered question

2021-12-28

Eliminate

θ and ϕ

between the following equations:

{\begin{matrix} \sin θ + \sin ϕ = x \\ \cos θ + \cos ϕ = y \\ \tan \frac{θ}{2} \tan \frac{ϕ}{2} = z \end{matrix}

What Ive

Answer & Explanation

Cassandra Ramirez

Beginner2021-12-29Added 30 answers

By sum to product and product to sum formulas we have

{\begin{matrix} \sin θ + \sin ϕ = 2 \sin (\frac{θ + ϕ}{2}) \cos (\frac{θ - ϕ}{2}) = x \\ \cos θ + \cos ϕ = 2 \cos (\frac{θ + ϕ}{2}) \cos (\frac{θ - ϕ}{2}) = y \\ \tan \frac{θ}{2} \tan \frac{ϕ}{2} = \frac{\cos (\frac{θ - ϕ}{2}) - \cos (\frac{θ + ϕ}{2})}{\cos (\frac{θ - ϕ}{2}) + \cos (\frac{θ + ϕ}{2})} = z \end{matrix} \Rightarrow {\begin{matrix} 2 a b = x \\ 2 c b = y \\ \frac{b - c}{b + c} = z \end{matrix}

and since

a^{2} + c^{2} = 1

we obtain

b = \pm \frac{12}{\sqrt{x^{2} + y^{2}}}

c = \pm \frac{y}{\sqrt{x^{2} + y^{2}}}

and then

z = \frac{\pm \frac{12}{\sqrt{x^{2} + y^{2}}} \mp \frac{y}{\sqrt{x^{2} + y^{2}}}}{\pm \frac{12}{\sqrt{x^{2} + y^{2}}} \pm \frac{y}{\sqrt{x^{2} + y^{2}}}} = \frac{x^{2} + y^{2} - 2 y}{x^{2} + y^{2} + 2 y}

Joseph Lewis

Beginner2021-12-30Added 43 answers

Write

u = \tan \frac{θ}{2}, v = \tan \frac{ϕ}{2}

. Then we get

{\begin{matrix} 2 (u + v) (1 + u v) = x (1 + u^{2}) (1 + v^{2}) \\ 2 - 2 u^{2} v^{2} = y (1 + u^{2}) (1 + v^{2}) \\ u v = z \end{matrix}

Now setting s=u+v and using the third relation, we obtain

{\begin{matrix} 2 s (1 + z) = x (1 - 2 z + z^{2} + s^{2}) \\ 2 (1 - z^{2}) = y (1 - 2 z + z^{2} + s^{2}) \end{matrix}

Dividing, we obtain the relation

s = \frac{x (1 - z)}{y}

, which youve

karton

Expert2022-01-08Added 613 answers

Hints
Writing $θ = 2 p, ϕ = 2 q$
We have
$\frac{\sin (p + q)}{x} = \frac{\cos (p + q)}{y} = \pm \frac{1}{\sqrt{x^{2} + y^{2}}}$
Using this one can find $\cos (p - q)$
Now use
$\frac{z - 1}{z + 1} = \dots = \frac{\cos (p + q)}{\cos (p - q)}$

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