Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Let's simplify each of these expressions step-by-step.
1. [tex]\( 12 \sin 17^{\circ} \cos 17^{\circ} \)[/tex]:
Using the double-angle identity [tex]\( \sin 2\theta = 2 \sin \theta \cos \theta \)[/tex], we have:
[tex]\[ 12 \sin 17^{\circ} \cos 17^{\circ} = 12 \left(\frac{1}{2} \sin 34^{\circ}\right) = 6 \sin 34^{\circ} \][/tex]
2. [tex]\( 2 \frac{2 \tan 30^{\circ}}{1 - \tan^2 30^{\circ}} \)[/tex]:
Using the double-angle identity for tangent [tex]\( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \)[/tex], we have:
[tex]\[ 2 \frac{2 \tan 30^{\circ}}{1 - \tan^2 30^{\circ}} = 2 \tan 60^{\circ} = 2 \sqrt{3} \][/tex]
3. [tex]\( 32 \cos^2 42^{\circ} - 1 \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 2 \cos^2 \theta - 1 \)[/tex], we have:
[tex]\[ 32 \cos^2 42^{\circ} - 1 = 16 (2 \cos^2 42^{\circ} - 1) = 16 \cos 84^{\circ} \][/tex]
4. [tex]\( 42 \sin \frac{1}{2} \theta \cos \frac{1}{2} \theta \)[/tex]:
Using the double-angle identity [tex]\( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \)[/tex], we have:
[tex]\[ 42 \sin \frac{1}{2} \theta \cos \frac{1}{2} \theta = 21 \sin \theta \][/tex]
5. [tex]\( 51 - 2 \sin^2 22.5^{\circ} \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 1 - 2 \sin^2 \theta \)[/tex], where [tex]\( 2 \times 22.5^\circ = 45^\circ \)[/tex], we have:
[tex]\[ 51 - 2 \sin^2 22.5^{\circ} = 51 - (1 - \cos 45^{\circ}) = 50 + \frac{\sqrt{2}}{2} \][/tex]
6. [tex]\( 7 \cos^2 15^{\circ} - \sin^2 15^{\circ} \)[/tex]:
Using the identity [tex]\( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \)[/tex], we have:
[tex]\[ 7 \cos^2 15^{\circ} - \sin^2 15^{\circ} = 6 \cos 30^{\circ} = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \][/tex]
7. [tex]\( 82 \sin 2A \cos 2A \)[/tex]:
Using the double-angle identity [tex]\( \sin 4\theta = 2 \sin 2\theta \cos 2\theta \)[/tex], we have:
[tex]\[ 82 \sin 2A \cos 2A = 41 \sin 4A \][/tex]
8. [tex]\( 6 \frac{2 \tan \frac{1}{2} \theta}{1 - \tan^2 \frac{1}{2} \theta} \)[/tex]:
Using the double-angle identity [tex]\( \tan \theta = \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}} \)[/tex], we have:
[tex]\[ 6 \frac{2 \tan \frac{1}{2} \theta}{1 - \tan^2 \frac{1}{2} \theta} = 6 \tan \theta \][/tex]
9. [tex]\( 101 - 2 \sin^2 3 \theta \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 1 - 2 \sin^2 \theta \)[/tex], we have:
[tex]\[ 101 - 2 \sin^2 3 \theta = 101 - (1 - \cos 6\theta) = 100 + \cos 6\theta \][/tex]
10. [tex]\( 11 \frac{\tan 2\theta}{1 - \tan^2 2\theta} \)[/tex]:
Using the identity [tex]\( \tan 4\theta = \frac{\tan 2\theta}{1 - \tan^2 2\theta} \)[/tex], we have:
[tex]\[ 11 \frac{\tan 2\theta}{1 - \tan^2 2\theta} = 11 \tan 4\theta \][/tex]
11. [tex]\( 2 \cos^2 \frac{1}{2} \theta - 1 \)[/tex]:
Using the identity [tex]\( \cos \theta = 2 \cos^2 \frac{\theta}{2} - 1 \)[/tex], we have:
[tex]\[ 2 \cos^2 \frac{1}{2} \theta - 1 = \cos \theta \][/tex]
12. [tex]\( 13 \frac{1 - \tan^2 20^{\circ}}{\tan 20^{\circ}} \)[/tex]:
Using the identity [tex]\( \cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan 20^\circ} \)[/tex], we have:
[tex]\[ 13 \frac{1 - \tan^2 20^{\circ}}{\tan 20^{\circ}} = 13 \cot 40^{\circ} \][/tex]
13. [tex]\( \sec \theta \csc \theta \)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \csc \theta = \frac{1}{\sin \theta} \][/tex]
[tex]\[ \sec \theta \csc \theta = \frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta} = \frac{1}{\sin \theta \cos \theta} \][/tex]
14. [tex]\( 12 \sin x \cos x \)[/tex]:
Using the double-angle identity [tex]\( \sin 2\theta = 2 \sin \theta \cos \theta \)[/tex], we have:
[tex]\[ 12 \sin x \cos x = 6 \sin 2x \][/tex]
15. [tex]\( 151 - 2 \sin^2 \frac{1}{2} \theta \)[/tex]:
Using the identity [tex]\( \cos \theta = 1 - 2 \sin^2 \frac{\theta}{2} \)[/tex], we have:
[tex]\[ 151 - 2 \sin^2 \frac{1}{2} \theta = 151 - (1 - \cos \theta) = 150 + \cos \theta \][/tex]
1. [tex]\( 12 \sin 17^{\circ} \cos 17^{\circ} \)[/tex]:
Using the double-angle identity [tex]\( \sin 2\theta = 2 \sin \theta \cos \theta \)[/tex], we have:
[tex]\[ 12 \sin 17^{\circ} \cos 17^{\circ} = 12 \left(\frac{1}{2} \sin 34^{\circ}\right) = 6 \sin 34^{\circ} \][/tex]
2. [tex]\( 2 \frac{2 \tan 30^{\circ}}{1 - \tan^2 30^{\circ}} \)[/tex]:
Using the double-angle identity for tangent [tex]\( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \)[/tex], we have:
[tex]\[ 2 \frac{2 \tan 30^{\circ}}{1 - \tan^2 30^{\circ}} = 2 \tan 60^{\circ} = 2 \sqrt{3} \][/tex]
3. [tex]\( 32 \cos^2 42^{\circ} - 1 \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 2 \cos^2 \theta - 1 \)[/tex], we have:
[tex]\[ 32 \cos^2 42^{\circ} - 1 = 16 (2 \cos^2 42^{\circ} - 1) = 16 \cos 84^{\circ} \][/tex]
4. [tex]\( 42 \sin \frac{1}{2} \theta \cos \frac{1}{2} \theta \)[/tex]:
Using the double-angle identity [tex]\( \sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \)[/tex], we have:
[tex]\[ 42 \sin \frac{1}{2} \theta \cos \frac{1}{2} \theta = 21 \sin \theta \][/tex]
5. [tex]\( 51 - 2 \sin^2 22.5^{\circ} \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 1 - 2 \sin^2 \theta \)[/tex], where [tex]\( 2 \times 22.5^\circ = 45^\circ \)[/tex], we have:
[tex]\[ 51 - 2 \sin^2 22.5^{\circ} = 51 - (1 - \cos 45^{\circ}) = 50 + \frac{\sqrt{2}}{2} \][/tex]
6. [tex]\( 7 \cos^2 15^{\circ} - \sin^2 15^{\circ} \)[/tex]:
Using the identity [tex]\( \cos 2\theta = \cos^2 \theta - \sin^2 \theta \)[/tex], we have:
[tex]\[ 7 \cos^2 15^{\circ} - \sin^2 15^{\circ} = 6 \cos 30^{\circ} = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \][/tex]
7. [tex]\( 82 \sin 2A \cos 2A \)[/tex]:
Using the double-angle identity [tex]\( \sin 4\theta = 2 \sin 2\theta \cos 2\theta \)[/tex], we have:
[tex]\[ 82 \sin 2A \cos 2A = 41 \sin 4A \][/tex]
8. [tex]\( 6 \frac{2 \tan \frac{1}{2} \theta}{1 - \tan^2 \frac{1}{2} \theta} \)[/tex]:
Using the double-angle identity [tex]\( \tan \theta = \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}} \)[/tex], we have:
[tex]\[ 6 \frac{2 \tan \frac{1}{2} \theta}{1 - \tan^2 \frac{1}{2} \theta} = 6 \tan \theta \][/tex]
9. [tex]\( 101 - 2 \sin^2 3 \theta \)[/tex]:
Using the identity [tex]\( \cos 2\theta = 1 - 2 \sin^2 \theta \)[/tex], we have:
[tex]\[ 101 - 2 \sin^2 3 \theta = 101 - (1 - \cos 6\theta) = 100 + \cos 6\theta \][/tex]
10. [tex]\( 11 \frac{\tan 2\theta}{1 - \tan^2 2\theta} \)[/tex]:
Using the identity [tex]\( \tan 4\theta = \frac{\tan 2\theta}{1 - \tan^2 2\theta} \)[/tex], we have:
[tex]\[ 11 \frac{\tan 2\theta}{1 - \tan^2 2\theta} = 11 \tan 4\theta \][/tex]
11. [tex]\( 2 \cos^2 \frac{1}{2} \theta - 1 \)[/tex]:
Using the identity [tex]\( \cos \theta = 2 \cos^2 \frac{\theta}{2} - 1 \)[/tex], we have:
[tex]\[ 2 \cos^2 \frac{1}{2} \theta - 1 = \cos \theta \][/tex]
12. [tex]\( 13 \frac{1 - \tan^2 20^{\circ}}{\tan 20^{\circ}} \)[/tex]:
Using the identity [tex]\( \cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan 20^\circ} \)[/tex], we have:
[tex]\[ 13 \frac{1 - \tan^2 20^{\circ}}{\tan 20^{\circ}} = 13 \cot 40^{\circ} \][/tex]
13. [tex]\( \sec \theta \csc \theta \)[/tex]:
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \csc \theta = \frac{1}{\sin \theta} \][/tex]
[tex]\[ \sec \theta \csc \theta = \frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta} = \frac{1}{\sin \theta \cos \theta} \][/tex]
14. [tex]\( 12 \sin x \cos x \)[/tex]:
Using the double-angle identity [tex]\( \sin 2\theta = 2 \sin \theta \cos \theta \)[/tex], we have:
[tex]\[ 12 \sin x \cos x = 6 \sin 2x \][/tex]
15. [tex]\( 151 - 2 \sin^2 \frac{1}{2} \theta \)[/tex]:
Using the identity [tex]\( \cos \theta = 1 - 2 \sin^2 \frac{\theta}{2} \)[/tex], we have:
[tex]\[ 151 - 2 \sin^2 \frac{1}{2} \theta = 151 - (1 - \cos \theta) = 150 + \cos \theta \][/tex]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
