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I need step by step help solving these problems (not just answers).

Find the exact values of sin 2θ cos 2θ and tan 2θ given that sec θ = -25/24 and θ is on the interval [tex] [ \frac{ \pi }{2}, \pi ][/tex]

Using the side lengths of a triangle ([tex] \frac{5}{13} [/tex]) find the exact values of double angles sin 2θ, cos 2θ, and tan 2θ

Using the side lengths of a triangle ([tex] \frac{7}{24} [/tex]) find the exact values of half angles sin θ/2, cos θ/2, and tan θ/2

Use power reduction principles to rewrite [tex] cos^{2} x sin^{2}x [/tex]

Solve [tex] cot^{2}x=-cotx [/tex] on the interval [[tex]0,2 \pi [/tex]]

Respuesta :

secθ=-25/24 so cosθ=-24/25, sinθ=7/25 (quadrant 2) [√1-(24/25)²=√(625-576)/625=√(49/625)=7/25]
sin2θ=2sinθcosθ=-2×7×24/625=-336/625=-0.5376.
cos2θ=1-2sin²θ=1-2×49/625=527/625=0.8432.
tan2θ=sin2θ/cos2θ=-336/527 (=-0.6376 approx.)

The length of the third side is 12 making the Pythagorean triangle 5-12-13 (5²+12²=13²).
Assuming sinθ=5/13, then cosθ=12/13. sin2θ=2sinθcosθ=120/169; cos2θ=1-2sin²θ=1-50/169=119/169.
tan2θ=120/119.

Pythagorean triangle is 7-24-25. sinθ=2sin(θ/2)cos(θ/2), cosθ=1-2sin²(θ/2).
sin²θ=4sin²(θ/2)cos²(θ/2)=4sin²(θ/2)(1-sin²(θ/2)).
49/625= 4sin²(θ/2)-4sin⁴(θ/2); 4sin⁴(θ/2)-4sin²(θ/2)+49/625=0.
sin⁴(θ/2)-sin²(θ/2)+49/2500=0=(sin²(θ/2)-49/50)(sin²(θ/2)-1/50).
cosθ=1-2sin²(θ/2); 24/25=1-2sin²(θ/2), sin²(θ/2)=1/50, sin(θ/2)=1/(5√2)=√2/10.
cos(θ/2)=√1-1/50=7√2/10; tan(θ/2)=1/7.

(sin(x)cos(x))²=sin²(2x)/4.

This can be written cot(x)(cot(x)+1)=0. So cot(x)=0, x=π/2, 3π/2; or cot(x)=-1=1/tan(x), x=3π/4, 7π/4.