+592 \(\sin\theta = \frac{3}{5}\)
\(\cos \theta = \frac{4}{5}\)
Hypotenuse of the second triangle is: \(\sqrt{24^2+7^2} = 25\)
\(\sin \alpha = \frac{7}{25}\)
\(\tan \alpha = \frac{7}{24}\)
This means:
\(\theta = \arcsin \frac{3}{5} ≈ 36.87\)
\(\alpha = \arcsin \frac{7}{25} ≈ 16.26\)
Now:
1. \(\sin 2\theta ≈ \sin 2(36.87) ≈ \sin73.74 = 0.96\)
2. \(\cos 2\theta ≈ \cos 2(36.87) ≈ \cos 73.74 = 0.28\)
3. \(\sin 2\alpha ≈ \sin 2(16.26) ≈ \sin 32.52 = 0.5376\)
4. \(\tan 2\alpha ≈ \tan 2(16.26) ≈ \tan 32.52 ≈ 0.6376\)
+37146 sin 2theta = 2 sin cos
= 2 ( 3/5)(4/5) = 24/25
cos 2theta = cos^2 - sin^2
= (4/5 )^2 - (3/5)^2 = 16/25 - 9/25 = 7/25
SImilar for alpha....use double angle identities ..... tan(2alpha) = 2 tan / (1- tan2 ) .
