Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.5 Text Book Back Questions and Answers, Notes.
Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5
Question 1.
Find the value of cos 2A , A lies in the first quadrant when
(i) cos A = \(\frac{15}{17}\)
Answer:
we know sin2 A + cos2 A = 1
sin2 A = 1 – cos2 A
Since A lies in the first quadrant, sin A is positive
∴ sin A = \(\frac{8}{17}\)
cos 2A = cos2 A – sin2 A
(ii) sin A = \(\frac{4}{5}\)
Answer:
we know sin2 A + cos2 A = 1
cos2 A = 1 – sin2A
Since A lies in the first quadrant, cos A is positive
∴ cos A = \(\frac{3}{5}\)
cos 2A = cos2 A – sin2 A
(iii) tan A = \(\frac{16}{63}\)
Answer:
Question 2.
If θ be an acute angle, find
(i) sin \(\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\), when sin θ = \(\frac{1}{25}\)
(ii) cos \(\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\), when sin θ = \(\frac{8}{9}\)
Answer:
(i) sin \(\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\), when sin θ = \(\frac{1}{25}\)
(ii) cos \(\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\), when sin θ = \(\frac{8}{9}\)
Question 3.
If cos θ =\(\frac{1}{2}\left(a+\frac{1}{a}\right)\), show that cos 3θ = \(\frac{1}{2}\left(a^{3}+\frac{1}{a^{3}}\right)\)
Answer:
Question 4.
Prove that
cos 5θ = 16 cos5θ – 20 cos3θ + 5 cos θ
Answer:
Question 5.
Prove that sin 4α = 4 tan α \(\frac{1-\tan ^{2} \alpha}{\left(1+\tan ^{2} \alpha\right)^{2}}\)
Answer:
Question 6.
If A + B = 45°, show that (1 + tan A) (1 + tan B) = 2
Answer:
Given A + B = 45°
tan(A + B) = tan 45°
tan A + tan B = 1 – tan A . tan B —— (1)
(1 + tan A)(1 + tan B) = 1 + tan B + tan A + tan A tan B
= 1 + (tan A + tan B) + tan A tan B
= 1 + 1 – tan A tan B + tan A tan B (By equation (1))
= 2
Question 7.
Prove that (1 + tan 1°) (1 + tan 2°) (1 + tan 3°) …….. (1 + tan 44°) is a multiple of 4.
Answer:
1 + tan 44° = 1 + tan (45° – 1°)
(1 + tan 1°)(1 + tan 44°) = 2
Similarly (1 + tan 2°) (1 + tan 43°) = 2
(1 + tan 3°) (1 + tan 42°) = 2
(1 + tan 22°) (1 + tan 23°) = 2
= (1 + tan 1°) (1 + tan 2°)… (1 + tan 44°) = 2 × 2 × … 22 times
It is a multiple of 4.
Question 8.
Prove that tan \(\left(\frac{\pi}{4}+\theta\right)\) – tan \(\left(\frac{\pi}{4}-\theta\right)\) = 2 tan 2θ
Answer:
Question 9.
Show that
Answer:
Question 10.
prove that (1 + sec 2θ) (1 + sec 4θ) …………….. (1 + sec 2nθ) = tan 2nθ . cot θ.
Answer:
Question 11.
Prove that
Answer: