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Elden [556K]
4 years ago
9

Solve:

="cos^2(2x)-sin^2(2x)=0" align="absmiddle" class="latex-formula"> Thanks
Mathematics
1 answer:
Aliun [14]4 years ago
7 0

Answer:

x = pi/8   + pi/2  *n                    x = 3pi/8   +   pi /2 *n

Step-by-step explanation:

cos ^2 ( 2x) - sin ^2 (2x) = 0

Substitute u = 2x

cos ^2 ( u) - sin ^2 (u) = 0

We know cos ^2(x)-sin ^2(x)=cos (2x)

cos ( 2u) =0

Replacing u with 2x

cos  (2 *2x) =0

cos (4x) =0

cos u =0 when u = pi/2 + 2 pi n and 3pi/2 + 2 pi n  where n is an integer

4x = pi/2+2 pi n                    4x = 3pi/2+2pi n

x = pi/8   + pi/2  *n                    x = 3pi/8   +   pi /2 *n

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<h2><u>Answer with explanation</u>:</h2>

Given: In Δ ABC and ΔAEC,

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AB=BC           [given]

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⇒ ΔADB ≅ΔCDB   [By SSS congruence rule]

⇒ ∠ADB ≅∠CDB       ...(i)           [Corresponding parts of congruent triangles are congruent]

Since AC is a straight line,

∠ADB+∠CDB = 180°   [Linear pair]

⇒∠ADB+∠ADB=180°   [from (i)]

⇒2 ∠ADB=180°  

⇒∠ADB=90°  =∠CDB

Also ∠ADB+∠ADE=180°   [Linear pair]

⇒∠ADE=180°-∠ADB  = 180°-90°

⇒∠ADE=90°, i.e. ∠ADE is a right triangle.

Similarly, ∠CDB+∠CDE=180°

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AD= CD   [given]

ED=ED   [Common]

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