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How 1 + Cos (2A) = 2 Cos²A ?

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How Cos (2A) = 2 Cos²A - 1 ? As we know that  ➡  Cos(2A) = Cos²A - Sin²A .......(1) Also Sin²A + Cos²A = 1  or Sin²A = 1 - Cos²A......[put in (1) eq.] ➡Cos (2A) = Cos²A - (1 - Cos²A)                 = Cos²A - 1 + Cos²A Hence,            Cos (2A) = 2 Cos²A - 1          or 1 + Cos(2A) = 2 Cos²A ➡ If angle is half i.e. θ = A then, Cos (A) = 2 Cos²(A/2) - 1       or 1 + Cos(A) = 2 Cos²(A/2)

How 1 - Cos(2A) = 2Sin²A?

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How 1 - Cos2A = 2 sin²A? As we know that, Cos(2A) = Cos²A - Sin²A.........(1) ........👉{ Cos(2A) = Cos²A - Sin²A } also, Sin²A + Cos²A = 1       or Cos²A = 1- Sin²A......(2) ...............👉{ How Sin²θ + Cos²θ = 1?? } put eq. (2) into eq.(1) Cos(2A) = 1 - Sin²A - Sin²A Hence, Cos(2A) = 1 - Sin²A also, 1 - Cos2A = 2 sin²A  If angle(ϴ) is halved, then, CosA = 1 - Sin²(A/2) or  1 - CosA = 2 sin²(A/2)

How Cos(2A) = Cos²A - Sin²A or Cos(A) = Cos²(A/2) - Sin²(A/2) ?

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Derivation of  Cos(2A) = Cos²A - Sin²A or Cos(A) = Cos²(A/2) - Sin²(A/2) ? As we know that, Cos(A+B) = CosA CosB - SinA SinB ........👉👉{ Cos(A+B) = CosA CosB - SinA SinB } put B= A  we get, ⇒Cos(A+A) = CosA CosA - SinA SinA ⇒Hence, Cos(2A) = Cos²A - Sin²A ⇒If angle will be halved i.e. θ = A Then,      ⇒ Cos(A) = Cos²(A/2) - Sin²(A/2)

How Sin(2A) = 2SinA CosA or Sin(A) = 2Sin(A/2) Cos(A/2) ?

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Derivation of   Sin(2A) = 2 SinA CosA or Sin(A) = 2Sin(A/2) Cos(A/2)? As we know that, ⇒Sin(A+B) = SinA CosB + CosA SinB 👉👉( Sin(A+B) = SinA CosB + CosA SinB ) put B= A then, ⇒Sin(A+A) = SinA CosA + CosA SinA    ⇒Sin(2A) = SinA CosA + SinA CosA Hence, Sin(2A) = 2SinA CosA ⇒If angle will be halfed i.e. θ = A ⇒Hence,        Sin(A) = 2Sin(A/2) Cos(A/2)

How tan(π/4-θ) = (1−tanθ)/(1+tanθ) ?

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Derivation of tan(π/4-θ) = (1−tanθ)/(1+tanθ) As, tan(π/4+θ) = (1+tanθ)/(1-tanθ) ...........👉👉 Derivation of tan(π/4+θ) = (1+tanθ)/(1-tanθ) {Put θ= -θ} tan(π/4-θ) = [1 + tan(-θ)] /[1- tan(-θ)]         { As tan(-θ) = -tanθ } Hence, tan(π/4-θ) = (1-tanθ)/(1+tanθ)

How tan(π/4+θ) = (1+tanθ)/(1-tanθ) ?

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Derivation of tan(π/4+θ) = (1+tanθ)/(1-tanθ) As,  tan(A+B) = (tanA - tanB)/(1- tanA tanB)        👉 Derivation of tan(A+B) = (tanA + tanB)/(1- tanA tanB) { Put A = π/4 and B= θ } tan(π/4+θ) = [tan(π/4) + tanθ] /[1- tan(π/4)tanθ] As tan(π/4) = 1.......(i.e. tan 45° = 1) Hence,  tan(π/4+θ)= (1+tanθ)/(1-tanθ)

How tan(A-B) = (tanA - tanB)/(1+ tanA tanB) ?

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Derivation of tan(A-B) = (tanA - tanB)/(1+ tanA tanB) 👉👉 Derivation of tan(A+B) = (tanA + tanB)/(1- tanA tanB) 👉👉 Derivation of Sin(A+B) = SinA CosB + CosA SinB 👉👉 Derivation of Cos(A+B) = CosA CosB - SinA SinB

How tan(A+B) = (tanA + tanB)/(1- tanA tanB) ?

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Derivation of tan(A+B) = (tanA + tanB)/(1- tanA tanB )  👉👉 Derivation of Sin(A+B) = SinA CosB + CosA SinB 👉👉 Derivation of Cos(A+B) = CosA CosB - SinA SinB

(Trigonometry Value Table) Values of Trigonometric ratios

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👉 These are the values of trigonometric Ratios:- 👆 Picture showing values of sinθ,cosθ,tanθ where θ is the some angle. From the table values of Cosecant (Cosec), Secant(sec) and Cotangent (Cot) can be determined. As,  Cosecθ = 1/Sinθ         Secθ = 1/cosθ and  Cotθ = 1/tanθ or cosθ/sinθ 👉 What is Trigonometry? 👉 What is a right angled triangle?

Trigonometric Ratios

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Trigonometric Ratios :- There are only six possible ratios of sides of a right angled triangle. In Trigonometry, different names are given to those ratios of right angled triangle w.r.t. angle ∠θ These are the ratios:- {1}. P/H :-  Sine (sin) {2}. B/H :- Cosine (Cos) {3}. P/B :- tangent (tan) {4}. H/P :- Cosecant (cosec) {5}. H/B :- Secant (Sec) {6}. B/P :- Cotangent (Cot) where P :- Perpendicular B:- Base H:- Hypotenuse source:  Wikipedia Some simple identities : ⇒tanϴ = 1/ cotϴ or sinϴ/ cosϴ ⇒Cosecϴ = 1/ sinϴ ⇒Secϴ  = 1/ cosϴ ⇒Cotϴ = 1 / tanϴ or cosϴ/ sinϴ 👉👉 What is Trigonometry? 👉👉 Trigonometry Table (Values of Trigonometric ratios) 👉👉 How Sin²θ + Cos²θ = 1?

What is Trigonometry?

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What is Trigonometry? Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. 👉👉 What is a Right angled triangle?

How Cos(A-B) = CosA CosB + SinA SinB ?

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Derivation of Cos(A-B) = CosA CosB + SinA SinB 👉👉 Derivation of Cos(A+B) = CosA CosB - SinA SinB

How Cos(A+B) = CosA CosB - SinA SinB ?

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Derivation of Cos(A+B) = CosA CosB - SinA SinB

How Sin(A-B) = SinA CosB- CosA SinB

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How Sin(A-B) = SinA CosB- CosA SinB 👉👉( Derivation of Sin(A+B) = SinA CosB + CosA SinB )

How Sin(A+B) = SinA CosB + CosA SinB ?

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Derivation of Sin(A+B) = SinA CosB + CosA SinB

How 1 + cot²θ = cosec²θ?

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Derivation of 1 + cot²θ = cosec²θ As we know that, Sin²θ + Cos²θ = 1........👉👉( Derivation of Sin²θ + Cos²θ = 1 )   Divide by sin²θ both sides , we get   Sin²θ/sin²θ + Cos²θ/sin²θ = 1/sin²θ {1/sin²θ = cosec²θ  and Cos²θ/sin²θ= Cot²}   Hence,       1 + cot²θ = cosec²θ

How 1 + tan²θ = sec²θ?

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Derivation of tan²θ + 1 = sec²θ As we know that,               Sin²θ + Cos²θ = 1  ........👉👉( Derivation of Sin²θ + Cos²θ = 1 ) Divide by Cos²θ both sides ,    we get, Sin²θ/Cos²θ + Cos²θ/Cos²θ = 1/Cos²θ Hence,    tan²θ + 1 = sec²θ

How Sin²θ + Cos²θ = 1?

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How   Sin²θ + Cos²θ = 1...?? According to Pythagoras theorem, "the square of hypotenuse(H) is equal to the sum of square of base(B) and perpendicular(P)" i.e.     H² = B² + P² (Dividing both sides by H²) we get,     H²/H² = B²/H² + P²/H²        1   = Sin²θ + Cos²θ                       ....(B/H  = sinθ and P/H = cosθ) Hence, Sin²θ + Cos²θ = 1 How Sin²θ + Cos²θ = 1?

What is a Right angled triangle?

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Right angled triangle :   Right angled triangle is a triangle in which one angle is a right angle (that is, a 90-degree angle). # The relation between the sides and angles of the right angled is the basis for trigonometry. # Pythagoras Theorem states that The square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of other two sides. 

Trigonometric values w.r.t. Quadrants (1st, 2nd, 3rd and 4th)

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👉Important Rules for trigonometric values in 1st ,2nd ,3rd and 4th Quadrant. {1}.    Quadrant 1st: (90°-θ) and (360°+θ) All Trigonometric values are Positive (+). {2}. Quadrant 2nd :  (90°+θ) and (180°-θ) Only values of sine (sin) and Cosecant (cosec) is positive (+) rest all Trigonometric values are negative (-). {3}. Quadrant 3rd:  (180°+θ) and (270°-θ) Only values of tangent (tan) and Cotangent (cot) is positive (+) rest all Trigonometric values are negative (-). {4} . Quadrant 4th :  (270°+θ) and (360° -θ) Only values of cosine (cos) and Secant (sec) is positive (+) rest all Trigonometric values are negative (-). 👇 The picture showing the values of all Trigonometric ratios in different quadrants :- # A Simple Way to remember A DD : All values positive S UGAR : Sine,Cosine are positive T O : Tan, cot are positive C OFEE : cos ,sec are positive