A "double angle" is just a special case of an angle sum : 2 A = A + A 2A = A + A 2 A = A + A sin ( X + Y ) \sin(X+Y) sin ( X + Y ) cos ( X + Y ) \cos(X+Y) cos ( X + Y ) X = Y = A X = Y = A X = Y = A one substitution away from the addition formulas.
WHY it matters: These formulas let us collapse sin 2 A \sin 2A sin 2 A sin A \sin A sin A cos A \cos A cos A
HOW we use them: put X = Y = A X = Y = A X = Y = A
sin 2 A = sin ( A + A ) = sin A cos A + cos A sin A \sin 2A = \sin(A+A) = \sin A\cos A + \cos A \sin A sin 2 A = sin ( A + A ) = sin A cos A + cos A sin A
The two terms are identical, so they add:
Why this step? sin A cos A + cos A sin A \sin A\cos A + \cos A\sin A sin A cos A + cos A sin A 2 sin A cos A 2\sin A\cos A 2 sin A cos A
cos 2 A = cos ( A + A ) = cos A cos A − sin A sin A = cos 2 A − sin 2 A \cos 2A = \cos(A+A) = \cos A\cos A - \sin A\sin A = \cos^2 A - \sin^2 A cos 2 A = cos ( A + A ) = cos A cos A − sin A sin A = cos 2 A − sin 2 A
Now use the Pythagorean identity = = sin 2 A + cos 2 A = 1 = = ==\sin^2 A + \cos^2 A = 1== == sin 2 A + cos 2 A = 1 ==
Replace sin 2 A = 1 − cos 2 A \sin^2 A = 1 - \cos^2 A sin 2 A = 1 − cos 2 A
cos 2 A = cos 2 A − ( 1 − cos 2 A ) = 2 cos 2 A − 1 \cos 2A = \cos^2 A - (1 - \cos^2 A) = 2\cos^2 A - 1 cos 2 A = cos 2 A − ( 1 − cos 2 A ) = 2 cos 2 A − 1
Replace cos 2 A = 1 − sin 2 A \cos^2 A = 1 - \sin^2 A cos 2 A = 1 − sin 2 A
cos 2 A = ( 1 − sin 2 A ) − sin 2 A = 1 − 2 sin 2 A \cos 2A = (1 - \sin^2 A) - \sin^2 A = 1 - 2\sin^2 A cos 2 A = ( 1 − sin 2 A ) − sin 2 A = 1 − 2 sin 2 A
Intuition Why THREE forms?Which form you pick depends on what you have. If a problem only gives you cos A \cos A cos A sin A \sin A sin A power-reduction rule sin 2 A = 1 − cos 2 A 2 \sin^2 A = \dfrac{1-\cos 2A}{2} sin 2 A = 2 1 − cos 2 A
tan 2 A = tan ( A + A ) = tan A + tan A 1 − tan A tan A \tan 2A = \tan(A+A) = \frac{\tan A + \tan A}{1 - \tan A\tan A} tan 2 A = tan ( A + A ) = 1 − t a n A t a n A t a n A + t a n A
Top: tan A + tan A = 2 tan A \tan A + \tan A = 2\tan A tan A + tan A = 2 tan A tan A ⋅ tan A = tan 2 A \tan A\cdot\tan A = \tan^2 A tan A ⋅ tan A = tan 2 A
Why this step? Same substitution X = Y = A X=Y=A X = Y = A tan X tan Y \tan X\tan Y tan X tan Y tan 2 A \tan^2 A tan 2 A Domain note: undefined when tan 2 A = 1 \tan^2 A = 1 tan 2 A = 1 A = 45 ∘ , 135 ∘ , … A = 45^\circ, 135^\circ,\dots A = 4 5 ∘ , 13 5 ∘ , … 2 A = 90 ∘ 2A = 90^\circ 2 A = 9 0 ∘ tan \tan tan
Worked example Example 1 — Givensin A \sin A sin A sin 2 A \sin 2A sin 2 A cos 2 A \cos 2A cos 2 A sin A = 3 5 \sin A = \tfrac{3}{5} sin A = 5 3 A A A
Step 1: Find cos A \cos A cos A A A A cos A = 1 − sin 2 A = 1 − 9 25 = 4 5 \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \tfrac{9}{25}} = \tfrac{4}{5} cos A = 1 − sin 2 A = 1 − 25 9 = 5 4 Why? Pythagoras gives the other side; acute ⇒ positive root.
Step 2: sin 2 A = 2 sin A cos A = 2 ⋅ 3 5 ⋅ 4 5 = 24 25 \sin 2A = 2\sin A\cos A = 2\cdot\tfrac35\cdot\tfrac45 = \tfrac{24}{25} sin 2 A = 2 sin A cos A = 2 ⋅ 5 3 ⋅ 5 4 = 25 24
Step 3: cos 2 A \cos 2A cos 2 A sin A \sin A sin A cos 2 A = 1 − 2 sin 2 A = 1 − 2 ⋅ 9 25 = 7 25 \cos 2A = 1 - 2\sin^2 A = 1 - 2\cdot\tfrac{9}{25} = \tfrac{7}{25} cos 2 A = 1 − 2 sin 2 A = 1 − 2 ⋅ 25 9 = 25 7 Why Form 3? It avoids re-using cos A \cos A cos A
Worked example Example 2 — Prove an identityProve sin 2 A 1 + cos 2 A = tan A \dfrac{\sin 2A}{1 + \cos 2A} = \tan A 1 + cos 2 A sin 2 A = tan A
Step 1: Numerator sin 2 A = 2 sin A cos A \sin 2A = 2\sin A\cos A sin 2 A = 2 sin A cos A Step 2: Denominator — pick the form that makes 1 + cos 2 A 1+\cos 2A 1 + cos 2 A 1 + cos 2 A = 1 + ( 2 cos 2 A − 1 ) = 2 cos 2 A 1 + \cos 2A = 1 + (2\cos^2 A - 1) = 2\cos^2 A 1 + cos 2 A = 1 + ( 2 cos 2 A − 1 ) = 2 cos 2 A Why Form 2? The + 1 +1 + 1 − 1 -1 − 1 Step 3: 2 sin A cos A 2 cos 2 A = sin A cos A = tan A \dfrac{2\sin A\cos A}{2\cos^2 A} = \dfrac{\sin A}{\cos A} = \tan A 2 cos 2 A 2 sin A cos A = cos A sin A = tan A
Worked example Example 3 — Forecast-then-verify withtan 2 A \tan 2A tan 2 A tan A = 1 2 \tan A = \tfrac12 tan A = 2 1 tan 2 A \tan 2A tan 2 A Forecast: A ≈ 26.6 ∘ A \approx 26.6^\circ A ≈ 26. 6 ∘ 2 A ≈ 53.1 ∘ 2A \approx 53.1^\circ 2 A ≈ 53. 1 ∘ tan 53.1 ∘ ≈ 1.33 \tan 53.1^\circ \approx 1.33 tan 53. 1 ∘ ≈ 1.33 4 3 \tfrac43 3 4 Compute: tan 2 A = 2 ⋅ 1 2 1 − ( 1 2 ) 2 = 1 1 − 1 4 = 1 3 4 = 4 3 \tan 2A = \dfrac{2\cdot\frac12}{1 - (\frac12)^2} = \dfrac{1}{1 - \frac14} = \dfrac{1}{\frac34} = \tfrac43 tan 2 A = 1 − ( 2 1 ) 2 2 ⋅ 2 1 = 1 − 4 1 1 = 4 3 1 = 3 4
sin 2 A = 2 sin A \sin 2A = 2\sin A sin 2 A = 2 sin A Why it feels right: "double the angle = double the sine" pattern-matches to linear scaling.
Why it's wrong: sin \sin sin not linear. Test A = 30 ∘ A=30^\circ A = 3 0 ∘ sin 60 ∘ = 0.866 \sin 60^\circ = 0.866 sin 6 0 ∘ = 0.866 2 sin 30 ∘ = 1 2\sin 30^\circ = 1 2 sin 3 0 ∘ = 1 Fix: sin 2 A = 2 sin A cos A \sin 2A = 2\sin A\cos A sin 2 A = 2 sin A cos A cos A \cos A cos A
Common mistake Picking the wrongcos 2 A \cos 2A cos 2 A Why it feels right: all three forms are equal, so "any" should work.
Why it bites: they're equal in value but not equal in convenience . Using Form 1 in Example 2 leaves cos 2 − sin 2 + 1 \cos^2 - \sin^2 + 1 cos 2 − sin 2 + 1 Fix: match the form to the data / the term you want to cancel.
cos A \cos A cos A sin A \sin A sin A Why it feels right: cos A = 1 − sin 2 A \cos A = \sqrt{1-\sin^2 A} cos A = 1 − sin 2 A Fix: the square root has ± \pm ± A A A A A A
What is sin 2 A \sin 2A sin 2 A sin A , cos A \sin A,\cos A sin A , cos A 2 sin A cos A 2\sin A\cos A 2 sin A cos A State the three forms of cos 2 A \cos 2A cos 2 A cos 2 A − sin 2 A \cos^2 A-\sin^2 A cos 2 A − sin 2 A ;
2 cos 2 A − 1 \;2\cos^2 A-1 2 cos 2 A − 1 ;
1 − 2 sin 2 A \;1-2\sin^2 A 1 − 2 sin 2 A What is tan 2 A \tan 2A tan 2 A 2 tan A 1 − tan 2 A \dfrac{2\tan A}{1-\tan^2 A} 1 − tan 2 A 2 tan A Which cos 2 A \cos 2A cos 2 A cos A \cos A cos A 2 cos 2 A − 1 2\cos^2 A-1 2 cos 2 A − 1 Which cos 2 A \cos 2A cos 2 A sin A \sin A sin A 1 − 2 sin 2 A 1-2\sin^2 A 1 − 2 sin 2 A From which double-angle result do you get sin 2 A = 1 − cos 2 A 2 \sin^2 A=\frac{1-\cos 2A}{2} sin 2 A = 2 1 − c o s 2 A Rearranging
cos 2 A = 1 − 2 sin 2 A \cos 2A=1-2\sin^2 A cos 2 A = 1 − 2 sin 2 A Where is tan 2 A \tan 2A tan 2 A When
tan 2 A = 1 \tan^2 A=1 tan 2 A = 1 , i.e.
A = 45 ∘ , 135 ∘ , … A=45^\circ,135^\circ,\dots A = 4 5 ∘ , 13 5 ∘ , … Trick to derive every double-angle formula? Set
X = Y = A X=Y=A X = Y = A in the addition formulas
Simplify 1 + cos 2 A 1+\cos 2A 1 + cos 2 A Simplify 1 − cos 2 A 1-\cos 2A 1 − cos 2 A
Recall Feynman: explain to a 12-year-oldImagine you know how to add two angles together to find their sine and cosine. A "double angle" is just adding an angle to itself . So instead of learning brand-new spells, you use the ones you already know and put the same number in twice. When you tidy up the leftovers, you get shortcut formulas. cos \cos cos sin 2 + cos 2 = 1 \sin^2 + \cos^2 = 1 sin 2 + cos 2 = 1 sin 2 \sin^2 sin 2 cos 2 \cos^2 cos 2
Sine: "2 sic " → 2 s in·c os.Cos three forms: "C an S wap S quares" — start with cos 2 − sin 2 \cos^2-\sin^2 cos 2 − sin 2 2 cos 2 − 1 2\cos^2-1 2 cos 2 − 1 1 − 2 sin 2 1-2\sin^2 1 − 2 sin 2 1 1 1 2 2 2 2c²−1 and 1−2s² .Tan: looks like the addition formula with everything doubled/squared: top 2t, bottom 1−t² .
Compound (Addition) Angle Formulas — the parent identities these come from.Pythagorean Identity — powers the three cos 2 A \cos 2A cos 2 A Power-Reduction / Half-Angle Formulas — rearrangements of cos 2 A \cos 2A cos 2 A Integration of sin²x and cos²x — direct application.Triple Angle Formulas — built by combining double + single angle.Weierstrass t = tan(A/2) Substitution — extends the tan 2 A \tan 2A tan 2 A
Addition formulas sin cos tan X+Y
Pythagorean identity sin^2+cos^2=1
tan 2A = 2tanA / 1 - tan^2A
Power-reduction sin^2A = 1-cos2A / 2
Intuition Hinglish mein samjhoDekho, "double angle" ka matlab hai simple: 2 A = A + A 2A = A + A 2 A = A + A X X X Y Y Y A A A sin 2 A = 2 sin A cos A \sin 2A = 2\sin A\cos A sin 2 A = 2 sin A cos A sin A cos A + cos A sin A \sin A\cos A + \cos A\sin A sin A cos A + cos A sin A
cos 2 A \cos 2A cos 2 A sin 2 + cos 2 = 1 \sin^2 + \cos^2 = 1 sin 2 + cos 2 = 1 cos 2 A − sin 2 A \cos^2 A - \sin^2 A cos 2 A − sin 2 A sin 2 \sin^2 sin 2 1 − cos 2 1-\cos^2 1 − cos 2 2 cos 2 A − 1 2\cos^2 A - 1 2 cos 2 A − 1 cos 2 \cos^2 cos 2 1 − 2 sin 2 A 1 - 2\sin^2 A 1 − 2 sin 2 A cos A \cos A cos A sin A \sin A sin A
tan 2 A \tan 2A tan 2 A A , A A,A A , A 2 tan A 2\tan A 2 tan A 1 − tan 2 A 1 - \tan^2 A 1 − tan 2 A A = 45 ∘ A = 45^\circ A = 4 5 ∘ 2 A = 90 ∘ 2A = 90^\circ 2 A = 9 0 ∘ tan 90 ∘ \tan 90^\circ tan 9 0 ∘
Sabse badi galti: mat sochna sin 2 A = 2 sin A \sin 2A = 2\sin A sin 2 A = 2 sin A sin \sin sin cos A \cos A cos A
