I need help with one task I'm stuck, PLEASE

[Johulus]

The task is following :

\(\displaystyle \dfrac{4\, \sin\left(8^{\circ}\right)\, \cdot\, \sin\left(52^{\circ}\right)\, \cdot\, \sin\left(428^{\circ}\right)}{4\, \cos^3\left(22^{\circ}\right)\, -\, 3\, \sin\left(68^{\circ}\right)}\, =\, \)

All angles are in degrees.
Task is to calculate the value of this expression without calculator. The result should be 1 but I don't know how to...get to that.

 

[Ishuda]

You will just have to play around and find relationships between the angles. For example
sin (a+360) =sin(a)
428 = 360 + 68
cos(90-a) = sin(a)
90 - 68 = 22

So now we have

\(\displaystyle \dfrac{4 \,\sin(8^{\circ}) \, \cdot\, \sin(52^{\circ}) \, \cdot\, \sin(360^{\circ}\, +\, 68^{\circ})}{4 \, \cos^3(90^{\circ}\, -\,68^{\circ})\, -\, 3 \,\sin(68^{\circ})}\, =\, \dfrac{4 \,\sin(8^{\circ}) \,\cdot\, \sin(52^{\circ})}{4 \,\sin^2(68^{\circ}) \,- \,3}\)

etc.

EDIT: Between the angles AND trig functions I meant to say. BTW: that 8, 52, and 68, looks nice for relationships with 60 degrees, i.e. 68 = 60 + 8 and 52 = 60 - 8.

 

[Johulus]

Yes, I know about all that conversions.... etc. but I don't know what to do then. I got to that last what you wrote but I can't figure out what to do with it :/..... I will try again later

Hoped someone would give me a step by step guide but anyway maybe it's better that I try a few more times :/

 

[Johulus]

BUSTED Thanks on your effort by the way!


\(\displaystyle \dfrac{4\, \sin\left(8^{\circ}\right)\, \cdot\, \sin\left(52^{\circ}\right)\, \cdot\, \sin\left(428^{\circ}\right)}{4\, \cos^3\left(22^{\circ}\right)\, -\, 3\, \sin\left(68^{\circ}\right)}\, =\, \dfrac{4\, \sin\left(8^{\circ}\right)\, \cdot\, \sin\left(52^{\circ}\right)\, \cdot\, \sin\left(68^{\circ}\right)}{\, \sin\left(68^{\circ}\right)\, (4\, \sin^2\left(68^{\circ}\right)\, -\, 3)}\, \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\, \dfrac{4\, \sin\left(8^{\circ}\right)\, \cdot\, \sin\left(52^{\circ}\right)}{4\, (\sin^2\left(68^{\circ}\right)\, -\, \dfrac{3}{4})}\, =\, \dfrac{\sin\left(8^{\circ}\right)\, \sin\left(52^{\circ}\right)}{\sin^2\left(68^{\circ}\right)\, -\, \sin^2\left(60^{\circ}\right)} \qquad \qquad \color{red}{\sin(\alpha)\, - \sin(\beta)\, =\, 2\, \cos(\dfrac{\alpha\, +\, \beta}{2})\, \sin(\dfrac{\alpha\, -\, \beta}{2})}\, \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\, \dfrac{\sin\left(8^{\circ}\right)\, \sin\left(52^{\circ}\right)}{(\sin\left(68^{\circ}\right)\, -\, \sin\left(60^{\circ}\right)\, )(\sin\left(68^{\circ}\right)\, +\, \sin\left(60^{\circ}\right)\, )}\qquad \qquad \color{red}{\sin(\alpha)\, +\, \sin(\beta)\, =\, 2\, \sin(\dfrac{\alpha\, +\, \beta}{2})\, \cos(\dfrac{\alpha\, -\, \beta}{2})} \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\, \dfrac{\sin\left(8^{\circ}\right)\, \sin\left(52^{\circ}\right)}{2\, \cos\left(64^{\circ}\right)\, \sin\left(4^{\circ}\right)\, 2\, \sin\left(64^{\circ}\right)\, \cos\left(4^{\circ}\right)} \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\, \dfrac{\sin\left(8^{\circ}\right)\, \sin\left(52^{\circ}\right)}{\sin\left(128^{\circ}\right)\, \sin\left(8^{\circ}\right)} \qquad \qquad \color{red}{\sin(2\, \alpha)\, =\, 2\, \sin(\alpha)\, \cos(\alpha)} \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\, \dfrac{\sin\left(8^{\circ}\right)\, \sin\left(52^{\circ}\right)}{\, \sin\left(52^{\circ}\right)\, \sin\left(8^{\circ}\right)}\, =\, 1\, \)