Question

Trigonometric formulae for CLASS 11th..

plz guys..I have been getting wrong answers since last three times.....

if u r not in 11th then kindly don't answer​

Answer 1

Answer:

Trigonometry Formulas :

1. sin(−θ) = −sin θ
2. cos(−θ) = cos θ
3. tan(−θ) = −tan θ
4. cosec(−θ) = −cosecθ
5. sec(−θ) = sec θ
6. cot(−θ) = −cot θ

Product to Sum Formulas :

1. sin x sin y = 1/2 [cos(x–y) − cos(x+y)]
2. cos x cos y = 1/2[cos(x–y) + cos(x+y)]
3. sin x cos y = 1/2[sin(x+y) + sin(x−y)]
4. cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

1. sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]
2. sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]
3. cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]
4. cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Identities :

1. sin²A + cos² A = 1
2. 1+tan²A = sec²A
3. 1+cot² A = cosec² A

Sign of Trigonometric Functions

in Different Quadrants :

Quadrants → I II III IV

1. Sin A + + – –
2. Cos A + – – +
3. Tan A + – + -
4. Cot A + – + –
5. Sec A. + – – +
6. Cosec A + + – –

Basic Trigonometric Formulas for Class 11

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

Based on the above addition formulas for sin and cos, we get the following below formulas:

• sin(π/2-A) = cos A
• cos(π/2-A) = sin A
• sin(π-A) = sin A
• cos(π-A) = -cos A
• sin(π+A)=-sin A
• cos(π+A)=-cos A
• sin(2π-A) = -sin A
• cos(2π-A) = cos A

If none of the angles A, B and (A ± B) is an odd multiple of π/2, then

• tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]
• tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

If none of the angles A, B and (A ± B) is a multiple of π, then

• cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]
• cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

Some additional formulas for sum and product of angles:

• cos(A+B) cos(A–B)=cos²A–sin2B=cos²B–sin²A
• sin(A+B) sin(A–B) = sin²A–sin²B=cos²B–cos²A
• sinA+sinB = 2 sin (A+B)/2 cos (A-B)/2

Formulas for twice of the angles:

• sin2A = 2sinA cosA = [2tan A /(1+tan²A)]
• cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]
• tan 2A = (2 tan A)/(1-tan²A)

Formulas for thrice of the angles:

sin3A = 3sinA – 4sin³A

cos3A = 4cos³A – 3cosA

tan3A = [3tanA–tan³A]/[1−3tan²A]

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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